Answer:
An independent samples t-test
Step-by-step explanation:
We have two different groups and we want to test if the scores are equal or not , so the best appropiate mthod would be an independent t test. Here we show the steps to conduct this test.
Data given and notation
[tex]\bar X_{1}[/tex] represent the mean for group men
[tex]\bar X_{2}[/tex] represent the mean for group women
Assuming these values for the remaining data:
[tex]s_{1}[/tex] represent the sample standard deviation for men
[tex]s_{2}[/tex] represent the sample standard deviation for women
[tex]n_{1}[/tex] sample size for the group men
[tex]n_{2}[/tex] sample size for the group women
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
Concepts and formulas to use
Suppose that we need to conduct a hypothesis in order to check if the mean are equal or not, the system of hypothesis would be:
H0:[tex]\mu_{1} = \mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
For this case is better apply a t test to compare means since we don't know the population deviations, and the statistic is given by:
[tex]z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
We just need to replace in formula (1) and find the calculated value.
Find the critical value
In order to find the critical value we need to take in count that we are conducting a two tailed test, and we need a significance level provided in order to find the critical region
Statistical decision
If our calculates value [tex]t_{calculated}>t_{critical}[/tex] or [tex]t_{calculated}<t_{critical}[/tex] we reject the null hypothesis. In other case we FAIL to reject the null hypothesis.
According to Inc, 79% of job seekers used social media in their job search in 2018. Many believe this number is inflated by the proportion of 22- to 30-year-old job seekers who use social media in their job search. Suppose a survey of 22- to 30-year-old job seekers showed that 314 of the 370 respondents use social media in their job search. In addition, 281 of the 370 respondents indicated they have electronically submitted a resume to an employer. (a) Conduct a hypothesis test to determine if the results of the survey justify concluding the proportion of 22- to 30-year-old job seekers who use social media in their job search exceeds the proportion of the population that use social media in their job search. Use α = 0.05. State the null and alternative hypothesis. (Enter != for ≠ as needed.)
Answer:
Null hypothesis:[tex]p\leq 0.79[/tex]
Alternative hypothesis:[tex]p > 0.79[/tex]
[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]
[tex]p_v =P(z>2.786)=0.00267[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.
Step-by-step explanation:
1) Data given and notation
n=750 represent the random sample taken
X=314 represent the respondents that use social media in their job search.
[tex]\hat p=\frac{314}{370}=0.849[/tex] estimated proportion of respondents that use social media in their job search
[tex]p_o=0.79[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:
Null hypothesis:[tex]p\leq 0.79[/tex]
Alternative hypothesis:[tex]p > 0.79[/tex]
When we conduct a proportion test we need to use the z statisticc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>2.786)=0.00267[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.
A manufacturer is interested in determining whether it can claim that the boxes of detergent it sells contain, on average, more than 500 grams of detergent. The firm selects a random sample of 100 boxes and records the amount of detergent (in grams) in each box. The data are provided in the file P09_02.xlsx.
a.Identify the null and alternative hypotheses for this situation.
b.Is there statistical support for the manufacturers claim
Answer:
a) H0=Weight of detergent=500gr (null hypotheses )
H1=Weight of detergent>500 gr. (alternative hypotheses )
b) X test.
Step-by-step explanation:
a)
The statement from which we are going to establish the assumptions is that supplied by the manufacturer, which says that the average weight of the detergent is 500 g. So:
H0=Weight of detergent=500gr (null hypotheses )
H1=Weight of detergent>500 gr. (alternative hypotheses )
b)
You are not providing the source you mention where the sample of detergent boxes taken is located. So I will explain in general terms how the problem should be posed.
For this case, we can use Z test because the sample we are taking is larger than 30.
The first thing we have to do is calculate Z, which is given by:
[tex]Z=\frac{(X-Xav)}{SD}[/tex]
Where:
Xav=Average of x.
SD=Standard deviation.
Then, we look for the table and if the value we have just calculated is greater than that of the table, then we must reject the null hypothesis.
Researchers published a study in which they considered the incidence among the elderly of various mental health conditions such as dementia, bi-polar disorder, obsessive compulsive disorder, delirium, and Alzheimer's disease. In the U.S., 45% of adults over 65 suffer from one or more of the conditions considered in the study. Calculate the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration. Give your answer accurate to three decimal places in decimal form. (Example: 0.398)
Probability =
Answer:
Probability = 0.100.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Eithey they have one or more of these conditions. Or they do not. This means that the binomial probability distribution is used to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 750, p = 0.45[/tex]
So
[tex]E(X) = 750*0.45 = 337.5[/tex]
[tex]\sqrt{Var(X)} = \sqrt{750*0.45*0.55} = 13.62[/tex]
Calculate the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration.
This is the pvalue of Z when [tex]X = 320[/tex]. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{320 - 337.5}{13.62}[/tex]
[tex]Z = -1.28[/tex]
[tex]Z = -1.28[/tex] has a pvalue of 0.100.
So
Probability = 0.100.
The test statistic of z=2.48 is obtained when testing the claim that p > 0.7.
(a) Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.
(b) Find the P-value.
(c) Using a significance level of alpha=.05, should we reject H0 or should we fail to reject H0?
Answer:
a) Right-tailed test
b) 0.006569
c) We reject [tex]H_0[/tex].
Step-by-step explanation:
We are given the following information in the question:
[tex]z_{\text{statistic}} = 2.48[/tex]
We are testing the claim that p > 0.7
Alpha, α = 0.05
a) We have to identify type of test.
Since we test for claim p > 0.7, it is a one-tailed(right) test.
b) The p-value can be calculated with the help of standard normal table.
[tex]P(z>2.48) = 1 - P(z<2.48) = 0.006569[/tex]
C) Since the p-value is less than the significance level, we fail to accept the null hypothesis and reject the null hypothesis.
Thus, we reject [tex]H_0[/tex].
The hypothesis test in question is right-tailed. The p-value can be obtained from a Z-table or statistical software and is probably less than 0.01. Given a significance level of alpha=.05, we should reject H0 if the p-value is less than alpha.
Explanation:This question involves the concepts of hypothesis testing, p-value, and significance level which are typically covered in a college-level statistics course.
(a) The test is right-tailed because we are testing the claim that p > 0.7. A right-tailed test looks at whether the test statistic is greater than the hypothesized population parameter.
(b) To find the p-value, you would need to look up the value of z=2.48 in a standard normal (Z) table or use statistical software. The p-value represents the probability that the observed difference occurred by chance assuming the null hypothesis is true. The exact value would depend on your table or software, but generally, a z-score of 2.48 would yield a very small p-value, probably less than 0.01.
(c) If the p-value is less than your significance level (alpha=.05), you would reject the null hypothesis (H0). That happens because the evidence suggests that it is highly unlikely (less than a 5% chance) that the observed difference occurred by chance.
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In 1997, 46% of Americans said they did not trust the media when it comes to reporting the news and fairly. In 2007 poll of 1010 adults nationwide, 525 stated they did not trust media. At the α=0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997
Answer:
We can conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997 (P-value=0.00014).
Step-by-step explanation:
We have to perform an hypothesis test on a proportion.
The null and alternative hypothesis are:
[tex]H_0: \pi=0.46\\\\H_1: \pi>0.46[/tex]
The significance level is α=0.05.
The standard deviation is estimated as:
[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.46(1-0.46)}{1010} }=0.0157[/tex]
The z value for this sample is
[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{525/1010-0.46-0.5/1010}{0.0157} =\frac{0.52-0.46-0.00}{0.0157}=\frac{0.06}{0.0157} =3.822[/tex]
The P-value for z=3.822 is P=0.00014.
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
We can conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.
The test statistic of z equals=2.17 is obtained when testing the claim that pnot equals≠0.2170.217. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of alphaαequals=0.010.01, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0?
Answer:
a) Two tailed test
Null hypothesis:[tex]p=0.217[/tex]
Alternative hypothesis:[tex]p \neq 0.217[/tex]
b) [tex]p_v =2*P(Z>2.17)=0.03[/tex]
c) If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
We Fail to reject the null hypothesis H0
Step-by-step explanation:
Data given and notation
n represent the random sample taken
X represent the outcomes desired in the sample
[tex]\hat p[/tex] estimated proportion of interest
[tex]p_o[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is 0.217 or no:
a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.
Two tailed test
Null hypothesis:[tex]p=0.217[/tex]
Alternative hypothesis:[tex]p \neq 0.217[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
For this case the calculated value is given z =2.17
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
b. Find the P-value
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(Z>2.17)=0.03[/tex]
c. Using a significance level of alphaαequals=0.01, should we reject Upper H 0 or should we fail to reject Upper H 0?
If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
We Fail to reject the null hypothesis H0
Zak is ordering custom T-shirts for his soccer team. Long-sleeved shirts cost $15 each and short-sleeved shirts cost $10 each. Jacob can spend at most $270 and he wants to order at least 20 shirts. Write the constraints using inequalities.
Question 1 options:
15x + 10y < 270
15x + 10y ≤ 270
15x + 10y ≥ 270
x + y ≥ 20
x + y ≤ 20
x > 0
x ≥ 0
y < 0
y ≥ 0
15x+10y≤270 and x+y≥20 are the constraints for this situation.
Step-by-step explanation:
Given,
Cost of long sleeved shirt = $15
Cost of short sleeved shirt = $10
Amount to spend = $270
Shirts to order = 20
Let,
Long sleeved shirts = x
Short sleeved shirts = y
At most means he can not spend more than $270, therefore,
15x+10y≤270
At least 20 means, he needs minimum 20 or more, therefore,
x+y≥20
15x+10y≤270 and x+y≥20 are the constraints for this situation.
Keywords: inequality, addition
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This is a GCSE maths question which i don’t understand.
The petrol consumption for David's journey, considering bounds, is between 6.5439 km/l and 6.8182 km/l.
To determine the petrol consumption (c) for David's journey, we use the formula c = d/p, where d is the distance (187 km) and p is the amount of petrol used (28 litres).
Firstly, we calculate the maximum and minimum possible values for c by considering the upper and lower bounds of d and p:
Maximum value of c:
Upper bound of distance (d) = 187 + 0.5 (half of the smallest decimal place in 187 km) = 187.5 km
Lower bound of petrol used (p) = 28 - 0.5 (half of the smallest decimal place in 28 litres) = 27.5 litres
Maximum c = (187.5 km) / (27.5 litres) = 6.8182 km/l
Minimum value of c:
Lower bound of distance (d) = 187 - 0.5 = 186.5 km
Upper bound of petrol used (p) = 28 + 0.5 = 28.5 litres
Minimum c = (186.5 km) / (28.5 litres) = 6.5439 km/l
Therefore, considering bounds, the petrol consumption (c) for David's journey is between 6.5439 km/l and 6.8182 km/l.
A researcher is conducting a chi-square test for independence to evaluate the relationship between gender and preference for three different designs for a new automobile. Each individual in a sample of n=30 males and n=30 females selects a favorite design from the three choices. If the researcher obtains a chi-square statistic of X^2 = 4.81, what is the appropriate statistical decision for the test?
O Fail to reject the null hypothesis with either α = 0.05 or α = 0.01
O There is not enough information to determine the appropriate decision.
O Reject the null hypothesis with α = 0.05 but not with α = 0. 01
O Reject the null hypothesis with either α = 0.05 or α = 0.01
Answer:
Step-by-step explanation:
A circle of fourths is generated by starting at any note and stepping upward by intervals of a fourth (five half-steps). By what factor is the frequency of a tone increased if it is raised by a fourth? How many fourths are required to complete the entire circle of fourths? How many octaves are covered in a complete circle of fourths?
Answer:
a) Q₀ × 1.335 cps
b) 12 fourths
c) 5 octaves
Step-by-step explanation:
The circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)
(a) By what factor is the frequency of a tone increased if it is raised by a fourth?
consider the following exponential growth formula :
Q = Q₀ × f °
Q = Q₀ × 1.05946 ⁿ
Substituting n = 5
Q = Q₀ × 1.05946 ⁿ
Q = Q₀ × 1.05946⁵
Q = Q₀ × 1.335 cps
Therefore, note that five half-step higher will be increased by factor 1.335
(b) How many fourths are required to complete the entire circle of fourths?
SOLUTION
In each step, number of half steps = 5
Total number of half steps in one octave = 12 half- steps
Therefore, total number of fourths required to complete the entire circle of fourths = 12 fourths
(c) Total number of half steps in one octave = 12 half-steps
Total half steps in complete circle of fourths
= 12×5
= 60 half-steps
Calculating number of octaves (dividing it by 12)
= 60/12
= 5 octaves
It is given that the circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)
Consider the exponential growth formula :[tex]Q = Q_0 \times f ^0\\Q = Q_0 \times 1.05946^n[/tex]
Put n = 5 in the above formula, we get
[tex]Q = Q_0 ( 1.05946 ^n)\\Q = Q_0 ( 1.05946 ^5)\\Q = Q_0 (1.335) cps[/tex]
So, by 1.335 factor the frequency of a tone increased if it is raised by a fourth.
In each step, number of half steps = 5Total number of half steps in one octave = 12 half- steps
So, total number of fourths required to complete the entire circle of fourths = 12 fourths
Total number of half steps in one octave = 12 half-stepsTotal half steps in complete circle of fourths will be 12×5 = 60 half-steps
Now, the number of octaves is [tex]\frac{60}{12} =5[/tex] octaves.
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For the Sequence 2,6,10,14. Write the recursive rule, and a explicit rule
Answer:
The recursive formula for given series is: [tex]a_{n} = a_{n-1} + 4[/tex]
The explicit formula is given by: [tex]a_{n} = 10 + 5(n-1)[/tex]
Step-by-step explanation:
Determining the Recursive Formula:
As the given sequence is: 2,6,10,14The difference d can be computed by taking the difference between the consecutive terms of the mentioned sequence.d = 6 -2 = 10 - 6 = 14 - 10 = 4
It is observed that the difference between the consecutive terms remains constant) with common difference d = 4We know that if the difference between the consecutive terms remains constant), then the series is in arithmetic series. The recursive formula is: [tex]a_{n} = a_{n-1} + d[/tex]So, the recursive formula for given series is: [tex]a_{n} = a_{n-1} + 4[/tex]Determining the Explicit Rule or Formula:
An explicit formula defines the nth term of the sequence, where n being the term's location. In other words, a sequence can be defined as a formula in terms of n. So,
First determine the sequence whether the sequence is in arithmetic. As we know that 2,6,10,14 is in arithmetic. Then find the common difference. Here, d = 6 -2 = 10 - 6 = 14 - 10 = 4Establishing an explicit formula by analyzing the pattern. i.e. adding first term to the product of d (common difference) and one less than the term numberHence, the explicit formula is given by:
[tex]a_{n} = a_{1} + d (n-1)[/tex]where,
aₙ is the nth term of the sequence
n is the term number
a₁ is the first term
d is the common difference
As the given sequence is: 2,6,10,14
a₁ = first term = 2
d = common difference = 6 - 2 = 4
Using explicit formula:
[tex]a_{n} = a_{1} + d (n-1)[/tex]
[tex]a_{n} = 10 + 5(n-1)[/tex]
[tex]a_{n} = 10 + 5n-5[/tex]
[tex]a_{n} = 5n+ 5[/tex]
Keywords: recursive formula, explicit formula, sequence
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Final answer :
The recursive rule for the sequence is Sn = Sn- 1 4 with S1 = 2, and the unequivocal rule is Sn = 4n- 2. Both rules help in determining the sequence's terms either by using the former term or directly calculating any term's value.
Explanation:
To determine the recursive rule and the unequivocal rule for the sequence 2, 6, 10, 14, we need to observe the pattern of the sequence and produce formulas that can induce the terms of the sequence. This sequence is computation because the difference between each term is constant.
Recursive Rule
A recursive rule for a sequence defines each term grounded on the former term. For the given sequence, each term after the first is attained by adding 4 to the former term. The recursive rule can be written as Sn = Sn- 1 4, for n> 1 where S1 = 2 is the first term.
unequivocal Rule
An unequivocal rule, also known as a unrestricted- form expression, defines the utmost term of a sequence as a function of n without demanding the former term. For this computation sequence, the unequivocal rule can be determined by using the formula Sn = a( n- 1) d, where' a' is the first term and'd' is the common difference. The unequivocal rule for our sequence is Sn = 2( n- 1) × 4 = 4n- 2.
An automobile dealer wants to see if there is a relationship between monthly sales and the interest rate. A random sample of 4 months was taken. The results of the sample are presented below. Monthly Sales (Y) Interest Rate (In Percent) (X) 22 9.2 20 7.6 10 10.4 45 5.3 a) Use the method of least squares to compute an estimated regression line. b) Obtain a measure of how well the estimated regression line fits the data.
Answer:
a) [tex]y=-6.254 x +75.064[/tex]
b) r =-0.932
The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.
Step-by-step explanation:
Assuming the following dataset:
Monthly Sales (Y) Interest Rate (X)
22 9.2
20 7.6
10 10.4
45 5.3
Part a
And we want a linear model on this way y=mx+b, where m represent the slope and b the intercept. In order to find the slope we have this formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
With these we can find the sums:
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=278.65-\frac{32.5^2}{4}=14.5875[/tex]
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=696.9-\frac{32.5*97}{4}=-91.225[/tex]
And the slope would be:
[tex]m=\frac{-91.225}{14.5875}=-6.254[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{32.5}{4}=8.125[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{97}{4}=24.25[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=24.25-(-6.254*8.125)=75.064[/tex]
So the line would be given by:
[tex]y=-6.254 x +75.064[/tex]
Part b
For this case we need to calculate the correlation coefficient given by:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
[tex]r=\frac{4(696.9)-(32.5)(97)}{\sqrt{[4(278.65) -(32.5)^2][4(3009) -(97)^2]}}=-0.937[/tex]
So then the correlation coefficient would be r =-0.932
The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.
You want to determine whether the amount of coffee that "coffee drinkers" consume on a weekly basis differs depending on whether the coffee drinker also smokes cigarettes. You seek 20 volunteers from your college campus to participate. Ten college students reported that they consume coffee but do not smoke cigarettes; these comprise the Coffee-Only group, or Group 1. Another ten college students reported that they consumed coffee and also smoked cigarettes; these comprise the Coffee + Cigarettes group, or Group 2. Then you ask them to monitor the frequency of 8 ounce cups of coffee they consumed over a seven day period. Which of the following are the correct statements of the null and alternate hypotheses, H0 and HA?
H0: µ1 > µ2 and HA: µ1 ≤ µ2
H0: µ1 ≤ µ2 and HA: µ1 > µ2
H0: µ1 < µ2 and HA: µ1 ≥ µ2
H0: µ1 ≥ µ2 and HA: µ1 < µ2
H0: µ1 = µ2 and HA: µ1 ≠ µ2
H0: µ1 ≠ µ2 and HA: µ1 = µ2
Answer:
The correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]
Step-by-step explanation:
Consider the provided information.
Ten college students reported that they consume coffee but do not smoke cigarettes; these comprise the Coffee-Only group, or Group 1. Another ten college students reported that they consumed coffee and also smoked cigarettes; these comprise the Coffee + Cigarettes group, or Group 2.
The null hypothesis tells the population parameter is equal to the claimed value.
If there is no statistical significance in the test then it is know as the null which is denoted by [tex]H_0[/tex], otherwise it is known as alternative hypothesis which denoted by [tex]H_a[/tex].
The amount of coffee that "coffee drinkers" consume on a weekly basis differs depending on whether the coffee drinker also smokes cigarettes.
Thus, the required hypothesis are: [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]
Therefore, the correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]
Please help. I do not understand how to solve this
Answer:
X=2
Step-by-step explain
4\div 10(12-3x)=3\div 10(12x-16)
when we do cross multiply then we get
40(12-3x)=30(12x-16)
we can divide by 10 both side and solve then we get
48-12x=36x-48
and after solving this equation we get
x=2
evaluate M2 + MNP if M=3,N= 4, and P=7
M^2 + MNP
(3)^2 + (3)(4)(7)
9 + 12(7)
9 + 84
93
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A restaurant owner is interested in the proportion of his customers who order dessert. He looks at 65 randomly selected receipts. Match the vocabulary word with its corresponding example.
The proportion of the 65 randomly selected customers who ordered dessert.
The list of the 65 Yes or No answers for whether each customer ordered dessert.
The 65 restaurant patrons whose receipts were observed by the owner.
The proportion of all customers who order dessert.
The answer: Yes or No to whether a customer ordered dessert.
All customers who come to the restaurant.
a. Data
b. Sample
c. Population
d. Statistic
e. Parameter
f. Variable
The context of a restaurant owner examining dessert orders involves several statistical concepts: Data is the list of Yes or No answers, the Sample is the 65 restaurant patrons, Population is all restaurant customers, the Statistic is the proportion of the sample who ordered dessert, the Parameter is the proportion of all customers who order dessert, and the Variable is whether an individual customer ordered dessert.
Explanation:The question involves the concepts of statistics in Mathematics, particularly focusing on how different terminologies are used. Here is the matching:
a. Data: This is the list of the 65 Yes or No answers for whether each customer ordered dessert.b. Sample: This applies to the 65 restaurant patrons whose receipts were observed by the owner.c. Population: 'All customers who come to the restaurant' is the complete set, or population in statistical terms.d. Statistic: The proportion of the 65 randomly selected customers who ordered dessert is a statistic because it provides an estimate of a particular characteristic, obtained from the sample.e. Parameter: The proportion of all customers who order dessert is the parameter. It's a numerical characteristic of the population.f. Variable: The answer: Yes or No to whether a customer ordered dessert is the variable in this context, because it can change for each customer or change over time.Learn more about Statistics Vocabulary here:https://brainly.com/question/33544588
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A pediatrician has administered a flu vaccine to 1,000 children in a local community. The average number of children who contract the flu after they get vaccinated is 4 cases per 1,000 children. What is the probability that none of the 1,000 children will contract the flu this season? Hint: Use a Poisson distribution with λ = 4. Group of answer choices
Answer:
1.83%
Step-by-step explanation:
For the expected value of 4 cases per 1000 children. We can use a formula for Poisson distribution to calculate the probability that none of the 1000 students would contract the flu:
[tex]P(X = 0) = \frac{\lambda^ke^{-\lambda}}{k!}[/tex]
[tex]P(X = 0) = \frac{4^0 e^{-4}}{0!}[/tex]
[tex]P(X = 0) = \frac{1}{e^4} = 0.0183 [/tex]
So the probability of this to happens is 1.83%
The probability that none of the 1,000 children will contract the flu after getting vaccinated, given the average number of flu cases after vaccination is approximately 0.0183 or 1.83%, using the Poisson distribution.
Explanation:In this question, we are required to use the Poisson distribution to find the probability that none of the 1,000 children will contract the flu this season, given that the average number of cases per 1,000 children after vaccination is 4 (λ = 4).
A Poisson probability can be calculated using the formula P(x; λ) = (e^-λ * λ^x) / x!, where x represents the actual number of successes that result from the experiment, e is the base of the natural logarithm approximated to 2.71828, and λ is the mean number of successes that occur in a specified region.
To find the probability that zero children contract the flu, we use x = 0 in the formula. The Poisson probability P(0; 4) = ((e^-4 * 4^0) / 0!) = e^-4 = 0.0183.
So, the probability that none of the 1,000 children will contract the flu after getting vaccinated is approximately 0.0183 or 1.83%.
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Listed below are annual data for various years. The data are weights (metric tons) of imported lemons and car crash fatality rates per 100,000 population. Construct a scatterplot, find the value of the linear correlation coefficient r, and find theP-value using α=0.05. Is there sufficient evidence to conclude that there is a linear correlation between lemon imports and crash fatality rates? Do the results suggest that imported lemons cause car fatalities?
Lemon_Imports_(x) Crash_Fatality_Rate_(y)
230 15.8
264 15.6
359 15.5
482 15.3
531 14.9
1. What are the null and alternative hypotheses?
2. Construct a scatterplot.
3. The linear correlation coefficient r is
4. The test statistic t is
5. The P-value is
Because the P-value is ____ than the significance level 0.05, there ____ sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.
Do the results suggest that imported lemons cause carfatalities?
A. The results suggest that an increase in imported lemons causes car fatality rates to remain the same.
B. The results do not suggest any cause-effect relationship between the two variables.
C. The results suggest that imported lemons cause car fatalities.
D. The results suggest that an increase in imported lemons causes in an increase in car fatality rates.
Answer:
Because the P-value is _(0.02) less than the significance level 0.05, there is sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.
C. The results suggest that imported lemons cause car fatalities.
Step-by-step explanation:
Hello!
The study variables are:
X₁: Weight of imported lemons.
X₂: Car crash fatality rate.
The objective is to test if the imported lemons affect the occurrence of car fatalities. To do so you are asked to use a linear correlation test.
I've made a Scatterplot with the given data, it is attached to the answer.
To be able to use the parametric linear correlation you can use the parametric test (Person) or the nonparametric test Spearman. For Person, you need your variables to have a bivariate normal distribution. Since one of the variables is a discrete variable (ratio of car crashes) and the sample is way too small to make an approximation to a normal distribution, the best test to use is Spearman's rank correlation.
This correlation coefficient (rs) takes values from -1 to 1
If rs = -1 this means that there is a negative correlation between the variables
If rs= 1 this means there is a positive correlation between the variables
If rs =0 then there is no correlation between the variables.
The hypothesis is:
H₀: There is no linear association between X₁ and X₂
H₁: There is a linear association between X₁ and X₂
α: 0.05
To calculate the Spearman's correlation coefficient you have to assign ranks to the observed values of each variable, from the smallest to the highest). Then you have to calculate the difference (d)between the ranks and the square of that difference (d²). (see attachment)
The formula for the correlation coefficient is:
[tex]rs= 1 - \frac{6* (sum of d^2)}{(n-1)n(n+1)}[/tex]
[tex]rs= 1 - \frac{6* (40)}{4*5*6}[/tex]
rs= -1
For this value of the correlation coefficient, the p-value is 0.02
Since the p-value (0.02) is less than the significance level (0.05) the decision is to reject the null hypothesis. In other words, there is a linear correlation between the imported lemons and the car crash fatality ration, this means that the modification in the lemon import will affect the car crash fatality ratio.
Note: the correlation coefficient is negative, so you could say that there is a correlation between the variables and this is negative (meaning that when the lemon import increases, the car crash fatality ratio decreases)
I hope it helps!
Nike has decided to sell special high quality insoles for its line of basketball tennis shoes. It has fixed costs of $6 million and unit variable costs of $5 per pair. Nike would like to earn a profit of $2 million; how many pairs must they sell at a price of $15? a. 100,000 kits b. 200,000 kits c. 600,000 kits d. 800,000 kits e. 1,400,000 kits
Answer:
option (d) 800,000
Step-by-step explanation:
Let the number of pairs that must be sold be 'x'
Thus,
Total variable cost for x units = $5x
Therefore,
Total cost = Fixed cost + Total variable cost
= $6 million + $5x
Now,
Profit = Cost - Revenue
Thus,
$2 million = $15x - ( $6 million + $5x )
or
$2 million + $6 million = $10x
or
$8 million = $10x
or
$8,000,000 = $10x
or
x = 800,000
Hence,
option (d) 800,000
A statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points. The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.
Answer:
There is no evidence that there is no significant difference between the sample means
Step-by-step explanation:
given that a statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points
The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.
Group One tailed X Two tailed Y
Mean 7.8100 7.6400
SD 1.0600 1.3300
SEM 0.1185 0.1487
N 80 80
[tex]H_0:\bar x=\bar y\\H_a: \bar x \neq \bar y[/tex]
(Two tailed test)
The mean of One tailed X minus Two tailed Y equals 0.1700
t = 0.8940
df = 158
p value =0.3727
p is greater than alpha 0.05
There is no evidence that there is no significant difference between the sample means
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.004.00 Mbps. Conduct the hypothesis test using these results. Use a 0.050.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. LOADING... Click the icon to view the display from technology. What are the null and alternative hypotheses? A. Upper H 0H0: muμequals=4.004.00 Mbps
Answer:
Null hypothesis:[tex]\mu \geq 4.0[/tex]
Alternative hypothesis:[tex]\mu < 4.00[/tex]
[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]
[tex]df=n-1=45-1=44[/tex]
Since is a left-sided test the p value would be:
[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=3.48[/tex] represent the sample mean
[tex]s=1.150075[/tex] represent the sample standard deviation
[tex]n=45[/tex] sample size
[tex]\mu_o =4.00[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 4.00 :
Null hypothesis:[tex]\mu \geq 4.0[/tex]
Alternative hypothesis:[tex]\mu < 4.00[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]
P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=45-1=44[/tex]
Since is a left-sided test the p value would be:
[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.
A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known that 2% of adults have neither a cell phone nor a landline. 3. What is the probability that an adult selected at random has both a landline and a cell phone? A. 0.58 B. 0.98 C. 0.6059 D. None of these Work: 4. Given an adult has a cell phone, what is the probability he does not have a landline?
A. 0.27
B. 0.25
C. 0.3012
D. None of these
Answer:
3. What is the probability that an adult selected at random has both a landline and a cell phone?
A. 0.58
4. Given an adult has a cell phone, what is the probability he does not have a landline?
C. 0.3012
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that an adult has a landline at his residence.
B is the probability that an adult has a cell phone.
C is the probability that a mean is neither of those.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that an adult has a landline but not a cell phone and [tex]A \cap B[/tex] is the probability that an adult has both of these things.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The sum of all the subsets is 1:
[tex]a + b + (A \cap B) + C = 1[/tex]
2% of adults have neither a cell phone nor a landline.
This means that [tex]C = 0.02[/tex].
73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)
So [tex]A = 0.73, B = 0.83[/tex].
What is the probability that an adult selected at random has both a landline and a cell phone?
This is [tex]A \cap B[/tex].
We have that [tex]A = 0.73[/tex]. So
[tex]A = a + (A \cap B)[/tex]
[tex]a = 0.73 - (A \cap B)[/tex]
By the same logic, we have that:
[tex]b = 0.83 - (A \cap B)[/tex].
So
[tex]a + b + (A \cap B) + C = 1[/tex]
[tex]0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1[/tex]
[tex](A \cap B) = 0.75 + 0.83 - 1 = 0.58[/tex]
So the answer for question 3 is A.
4. Given an adult has a cell phone, what is the probability he does not have a landline?
83% of the adults have a cellphone.
We have that
[tex]b = B - (A \cap B) = 0.83 - 0.58 = 0.25[/tex]
25% of those do not have a landline.
So [tex]P = \frac{0.25}{0.83} = 0.3012[/tex]
The answer for question 4 is C.
Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the probability that a randomly selected woman has a foot length less than 10.0 in.b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.
Answer:
a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 9.6, \sigma = 0.5[/tex].
a. Find the probability that a randomly selected woman has a foot length less than 10.0 in
This probability is the pvalue of Z when [tex]X = 10[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10 - 9.6}{0.5}[/tex]
[tex]Z = 0.8[/tex]
[tex]Z = 0.8[/tex] has a pvalue of 0.7881.
So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.
When X = 10, Z has a pvalue of 0.7881.
For X = 8:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8 - 9.6}{0.5}[/tex]
[tex]Z = -3.2[/tex]
[tex]Z = -3.2[/tex] has a pvalue of 0.0007.
So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.
Now we have [tex]n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1[/tex].
This probability is 1 subtracted by the pvalue of Z when [tex]X = 9.8[/tex]. So:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9.8 - 9.6}{0.1}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.
Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity.(a) Solve this differential equation for c=0.1, K=2000, and initial population P0=500. P(t)= .(b) Compute the limiting value of the size of the population. limt→[infinity]P(t)= .(c) At what value of P does P grow fastest? P= .
Answer:
A) [tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]
B) P(t→∞)=2000
C) [tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]
Step-by-step explanation:
Given differential eq is
[tex]\frac{dP}{dt}=c ln (\frac{K}{P})P[/tex] --- (1)
Eq is separable
[tex]\frac{1}{ln (\frac{K}{P})P}dP=cdt[/tex] --- (2)
[tex]let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du[/tex]
substituting in (2)
[tex]-\frac{du}{u}=dt[/tex]
Integrating both sides
[tex]\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B\\[/tex]
Back substituting value of u
[tex]ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}\\[/tex]---(3)
at t =0
[tex]ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|[/tex]
from (3)
[tex]ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}[/tex]
[tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]
B) [tex]\lim{t \to \infty}[/tex]
[tex]P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000\\[/tex]
which is the carrying capacity.
C) To find the fastest growth rate we have to maximize [tex]\frac{dP}{dt}[/tex]
From given differential eq
[tex]\frac{dP}{dt}=cln|\frac{K}{P}|P[/tex]
so function to maximize is
[tex]f(P)=cln|\frac{K}{P}|P[/tex]
[tex]f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P[/tex]
[tex]f'(P)=c[ln|\frac{K}{P}|-1][/tex]
To maximize find f'(P)=0
[tex]c[ln|\frac{K}{P}|-1]=0[/tex]
[tex]ln|\frac{K}{P}|=1[/tex]
[tex]\frac{K}{P}=e[/tex]
[tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]
The Gompertz function models population growth considering the carrying capacity. solve the Gompertz differential equation for c=0.1, K=2000, P0=500. The carrying capacity (K) gives the limiting value of the population size (2000). The time at which the population growth is fastest can be calculated by taking the derivative of the population function, setting it to zero and solving for P.
Explanation:The Gompertz function is a model of growth that was developed to model population growth, considering the factor of carrying capacity. Instead of compound exponential growth, it models the growth as slowing down as it reaches the limit of the carrying capacity. In your particular case, solving the differential equation for the variables provided (c=0.1, K=2000, P0=500) would require the integration techniques and use of logarithmic functions.
(a): The full solution for the Gompertz function involves advanced mathematics, you should expect some intricate function in the form of P(t) = ... The particulars depend on the specifics of the integration process.
(b): The limiting value of the population size as t→infinity (limt→[infinity]P(t) will be K, which in this instance equals 2000. This is due to the concept of carrying capacity. Beyond this value, the environment/conditions can no longer support additional growth.
(c): Finding the time at which population growth is fastest involves setting the derivative of the population function to zero and solving for P. The solution P=... can be calculated using standard techniques of calculus.
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Allen runs at an average rate of 9 mi/hr and walks at a average rate of 3 mi/hr. Write an equation in standard form to relate the times he can spend walking and running if he travels 30 miles. If he walks for 4 hours, for how long will he run?
Answer:
The equation would be [tex]30=12+9x[/tex].
Allen will Run for 2 hours.
Step-by-step explanation:
Given:
Total Distance traveled = 30 miles
Average rate of walking = 3 mi/hr
Average rate of running = 9 mi/hr
Time for walking = 4 hours
W need to find the time for running.
Solution:
Let Number of hrs required for running be 'x'.
Total Distance traveled is equal Distance Traveled in walking plus Distance traveled in Running
But Distance is equal Rate multiplied by Time.
Framing in equation form we get;
Hence Total Distance Traveled = Rate of walking × Hours of walking + Rate of Running × Hours of running.
Substituting the values we get;
[tex]30 = 3\times4 +9x\\\\30=12+9x[/tex]
Hence The equation would be [tex]30=12+9x[/tex]
Solving above equation we get;
[tex]9x=30-12\\\\9x=18\\\\x=\frac{18}{9}=2\ hrs[/tex]
Hence Allen will Run for 2 hours.
find the number of subsets of s = 1 2 3 . . . 10 that contain exactly 4 elements including 3 or 4 but not both.
Answer:
112
Step-by-step explanation:
Let A be a subset of S that satisfies such condition.
If 3∈A, then the other three elements of A must be chosen from the set B={1,2,5,6,7,8,9,10} (because 3 cannot be chosen again and 4 can't be alongside 3). B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A (the binomial coefficient counts this). The remaining elements determine A uniquely, then there are 56 subsets A.
If 4∉A, we have to choose the remaining elements of A from the set B={1,2,5,6,7,8,9,10}. B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A. Thus, there are 56 choices for A.
By the sum rule, the total number of subsets is 56+56=112
Nathan and Carl are running for the mayor of Middletown, in which 60% of the voters favor Nathan and 40% support Carl. A poll is conducted in which 100 residents, selected at random, are asked their preference. What is the likelihood that the poll will show a majority in favor of Carl?
A. 0.3409
B. 0.0068
C. 0.0207
D. 0.1976
Answer:
C. 0.0207
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they vote for Carl, or they vote for Nathan. So we use the binomial probability distribution.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]\mu = 100*0.4 = 40[/tex]
[tex]\sigma = \sqrt{100*0.4*0.6} = 4.9[/tex]
What is the likelihood that the poll will show a majority in favor of Carl?
This is 1 subtracted by the pvalue of Z when X = 50. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{50 - 40}{4.9}[/tex]
[tex]Z = 2.04[/tex]
[tex]Z = 2.04[/tex] has a pvalue of 0.9793. So the answer is 1-0.9793 = 0.0207.
The table below gives the number of hours seven randomly selected students spent studying and their corresponding midterm exam grades. Using this data, consider the equation of the regression line, yˆ = b0 + b1x, for predicting the midterm exam grade that a student will earn based on the number of hours spent studying. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.
Hours Studying Midterm Grades
1.5 67
2.5 69
3 80
3.5 81
4 86
4.5 89
5.5 94
Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.
Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.
Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.
Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.
Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.
Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places
Answer:
1) [tex]m=\frac{77}{10.5}=7.333[/tex]
2) [tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]
3) [tex]\hat y=7.333(0)+55.190=55.190[/tex]
4) False. The values predited will fall on the same line since we are estimating the values with just one line.
5) [tex]\hat y=7.333(4.5)+55.190=88.190[/tex]
6) [tex]R^2 = (0.971^2) =0.943[/tex]
And that means that the linear model explains 94.29% of the variation.
Step-by-step explanation:
We assume that the data is this one:
x:1.5,2.5, 3, 3.5 , 4, 4.5 ,5.5
y: 67, 69, 80, 81, 86, 89, 94.
Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
So we can find the sums like this:
[tex]\sum_{i=1}^n x_i = 1.5+2.5+3+3.5+4+4.5+5.5=24.5[/tex]
[tex]\sum_{i=1}^n y_i =67+ 69+ 80+ 81+ 86+ 89+ 94=566[/tex]
[tex]\sum_{i=1}^n x^2_i =1.5^2+2.5^2+3^2+3.5^2+4^2+4.5^2+5.5^2=96.25[/tex]
[tex]\sum_{i=1}^n y^2_i =67^2+69^2+80^2+81^2+86^2+89^2+94^2=46364[/tex]
[tex]\sum_{i=1}^n x_i y_i =1.5*67+2.5*69+3*80+3.5*81+4*86+4.5*89+5.5*94=2058[/tex]
With these we can find the sums:
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=96.25-\frac{24.5^2}{7}=10.5[/tex]
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=2058-\frac{24.5*566}{7}=77[/tex]
And the slope would be:
[tex]m=\frac{77}{10.5}=7.333[/tex]
Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.
Now we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{24.5}{7}=3.5[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{566}{7}=80.857[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]
So the line would be given by:
[tex]\hat y=7.333 x +55.190[/tex]
Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.
[tex]\hat y=7.333(0)+55.190=55.190[/tex]
Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.
False. The values predited will fall on the same line since we are estimating the values with just one line.
Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.
[tex]\hat y=7.333(4.5)+55.190=88.190[/tex]
Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places
n=7 [tex] \sum x = 24.5, \sum y = 566, \sum xy =2058, \sum x^2 =96.25, \sum y^2 =46364[/tex]
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
[tex]r=\frac{7(2058)-(24.5)(566)}{\sqrt{[7(96.25) -(24.5)^2][7(46364) -(566)^2]}}=0.971[/tex]
And the determination coeffcient is just the square of the correlation coefficient given by:
[tex]R^2 = (0.971^2) =0.943[/tex]
And that means that the linear model explains 94.3% of the variation.
Regression analysis can be done to find the relationship between hours spent studying and midterm exam grades using the regression line equation yˆ = b0 + b1x. But doing this requires statistical computations and ensuring the correlation coefficient is statistically significant. The coefficient of determination, which indicates the proportion of the variance in the dependent variable predictable from the independent variable, is also needed.
Explanation:The equation of the regression line you're referring to is written as yˆ = b0 + b1x, where yˆ is the predicted value of the dependent variable (in this case, Midterm Grades), b0 is the y-intercept, b1 is the slope or regression coefficient, and x is the value of the independent variable (Hours Studying). We can use the given data to find the estimated slope and y-intercept, value of yˆ at x = 0, test the truthfulness of the statement, and find the estimated value of y when x = 4.5.
However, being able to do this requires the use of statistical software or methods to calculate corresponding values. Also, this also requires us to know the value of the correlation coefficient to determine if it's statistically significant, as it's not wise to make predictions using a regression line with a correlation coefficient that's not statistically significant.
The last piece to find is the value of the coefficient of determination, which shows the proportion of the dependent variable's variance that's predictable from the independent variable(s). This value also needs computation using formulas or statistical software.
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An educator evaluates the effects of small, medium, and large class sizes on academic performance among male and female students. Identify each factor and the levels of each factor in this example. (Select all that apply.)
a. gender (two levels: male, female)
b. academic performance (three levels: above average, average, below average) class size (two levels: small, large)
c. academic performance (two levels: passing, failing)
d. class size (three levels: small, medium, large)
e. gender (three levels: male, female, trans)
Answer:
a. gender (two levels: male, female)
d. class size (three levels: small, medium, large)
Step-by-step explanation:
The study is going to be made to identify how male and female students react to the class size(small, medium, large). So there are two correct options, since the gender and the class size are factors in this research.
The correct options are:
a. gender (two levels: male, female)
d. class size (three levels: small, medium, large)
In this example, the factors and levels are gender, academic performance, and class size.
Explanation:The factors and levels in this example are:
a. Gender (two levels: male, female)
b. Academic performance (three levels: above average, average, below average)
c. Class size (two levels: small, large)
Therefore, the answer is a, b, and c.
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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444.0 gram setting. It is believed that the machine is underfilling the bags. A 40 bag sample had a mean of 443.0 grams. A level of significance of 0.02 will be used. Determine the decision rule. Assume the standard deviation is known to be 23.0.
Answer:
We conclude that the bag filling machine works correctly at the 444.0 gram setting.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 444.0 gram
Sample mean, [tex]\bar{x}[/tex] = 443.0 grams
Sample size, n = 40
Alpha, α = 0.02
Population standard deviation, σ = 23.0 grams
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 444.0\text{ grams}\\H_A: \mu < 444.0\text{ grams}[/tex]
We use one-tailed(left) z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{443 - 444}{\frac{23}{\sqrt{40}} } =-0.274[/tex]
Now, [tex]z_{critical} \text{ at 0.02 level of significance } = -2.054[/tex]
Since,
[tex]z_{stat} < z_{critical}[/tex]
We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the bag filling machine works correctly at the 444.0 gram setting.
The decision rule for the hypothesis test to check machine calibration involves finding the z-value for a sample mean of 443.0 grams and comparing it with the critical z-value corresponding to a 0.02 significance level for a left-tailed test. If the z-value is less than the critical value, the null hypothesis is rejected.
Explanation:To determine the decision rule for testing whether the bag filling machine is correctly set at 444.0 grams when we suspect underfilling, we need to perform a hypothesis test using the z-test since the standard deviation is known. Given the sample size (n = 40), sample mean (μ = 443.0 grams), population mean (μ0 = 444.0 grams), population standard deviation (σ = 23.0 grams), and a level of significance of 0.02, we can formulate the null hypothesis (H0: μ = μ0) and the alternative hypothesis (H1: μ < μ0).
The decision rule involves comparing the computed z-value to the critical value from the standard normal distribution at the 0.02 level of significance for a left-tailed test. If the computed z-value is less than the critical z-value, we reject the null hypothesis and accept that the machine is underfilling. The critical z-value is found using the z-table, which corresponds to a cumulative probability of 0.02.