Answer:
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
Answer:
1) 100 - 40 = 60
2) 800000 - 200000 = 600000
3) 300 + 80 = 410
4) 3000 + 800 = 1100
5) 600× 800 = 480000
6) 80×200 = 16000
7) 700×900 = 630000
8) 100×1000 = 100000
9) 800× 500 = 400000
10) 4000×80 = 320000
11) 800×100 = 80000
12) 800/20 = 40
13) 4000/70 = 57.1 = 60
14) 200/60 = 33.3 = 30
15) 600/80 = 7.5 = 8
16) 3000/200 = 15
17) 10000/200 = 50
18) 7000/300 = 23.3 = 20
19) 7000/400 = 17.5 = 18
20) 600000 / 2000 = 300
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to determine if the average age of all the students at the university is significantly different from 24. Assume the distribution of the population of ages is normal with a standard deviation of 2 years. At a .05 level of significance, it can be concluded that the mean age is _____.
a. not significantly different from 24
b. significantly different from 24
c. significantly less than 24
d. significantly less than 25
Answer:
Option b) significantly different from 24
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 24
Sample mean, [tex]\bar{x}[/tex] = 25
Sample size, n = 25
Alpha, α = 0.05
Population standard deviation, σ = 2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]
We use Two-tailed 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{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]
Now, we calculate the p-value from the standard normal table.
P-value = 0.0455
Since the p-value is less than the significance level, we fail to accept the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that mean age is significantly different from 24.
At 5% level of significance, it can be concluded that the mean age is B. Significantly different from 24.
How to explain the confidence level?From the information given, the university has had an average age of 25 years. The critical value at 5% level with the degree of freedom is 1.753 while the test statistic will be:
= (25 - 24)/2 × ✓26
= 1/2 × 4 = 2
Since the test statistic is greater than 1.753, one should reject the null hypothesis. Therefore, it shows that that the mean age is significantly different from 24.
Learn more about significance level on:
https://brainly.com/question/4599596
Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. In a left-tailed test at α = .05, which is the most accurate statement?
We would strongly reject the claim.
We would clearly fail to reject the claim.
We would face a rather close decision.
We would switch to α = .01 for a more powerful test.
Not sure what they mean. Can you please explain why it is that answer please?
Answer:
[tex]p_v =P(t_{17}<-1.728)=0.051[/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 FAIL to reject the null hypothesis.
The best option for this case would be:
We would switch to α = .01 for a more powerful test.
Step-by-step explanation:
Previous concepts and data given
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=37.8[/tex] represent the sample mean
[tex]s=5.4[/tex] represent the sample standard deviation
n=18 represent the sample selected
[tex]\alpha=0.05[/tex] significance level
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{37.8-40}{\frac{5.4}{\sqrt{18}}}=-1.728[/tex]
P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=18-1=17[/tex]
Since is a left tailed test the p value would be:
[tex]p_v =P(t_{17}<-1.728)=0.051[/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 FAIL to reject the null hypothesis.
The best option for this case would be:
We would switch to α = .01 for a more powerful test.
Since the p values is just a little higher than th significance level 0.051>0.05 but the values are two close. If we change the value of the significance by 0.01, we have that 0.051>0.01 and it's a more powerful test.
A quasi-experiment was conducted to compare men's and women's attitudes about extramarital affairs. Men and women who are married were recruited to complete a survey about their attitudes. The researchers then compared scores on the survey for men and women. The appropriate statistical test to analyze the data in this study is ______
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.
A geologist manages a large museum collection of minerals, whose mass (in grams) is known to be normally distributed. She knows that 60% of the minerals have mass less than 5000 g, and needs to select a random sample of n = 16 specimens for an experiment. With what probability will their average mass be less than 5000 g?
Answer:
[tex]P(\bar X < 5000)=P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Let X the random variable that represent the mass of minerals of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
2) Solution to the problem
For this case we know this condition given :
[tex]P(X<5000)=0.6[/tex]
We can use the Z score given by this formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
And using this formula we got:
[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]
And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"
So then we can do this:
[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex] (1)
By the other hand when we find the z score for the sample mean we have this:
[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And we want to find this probability:
[tex]P(\bar X < 5000)[/tex]
And if we use the z score formula we got:
[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex] (2)
And replacing condition (1) into equation (2) we got:
[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]
And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"
Use the data from problem:
52.2 43.8 50.3 51.1 48.3 47.8 48.3 47.4 50.1 50.5 51.4 54.2 54.4 48.6 54.5 47.3 50.3 48.1 46.6 50.2 50.5 48.2 46.3 48.1 49.4 50.5 47.7 50.1 45.6 49.3 44.4 47.2 47.6 56.9 48.9 49.9 46.3 44.9 51.2 48.5 49.2 46.6 47.3 45.3 49.2 51.1 49.2 50.0 49.8 48.2 47.2 42.6 46.9 46.5 47.3 46.5 47.7 49.2 46.3 48.5 53.4 48.0 50.0 49.7 48.8 48.3 48.7 48.1 48.2 48.6 48.3 48.3 48.3 48.3 48.6 48.2 48.3 48.7 48.1 48.5
a. Calculate the sample mean, sample median, sample variance, and sample standard deviation.
b. Construct a Stem and Leaf Plot, Histogram, and Box Plot.
Answer:
a) Sample mean: 48.71
Sample median: 48.3
Sample variance: 5.41
Sample standard deviation: 2.37
Step-by-step explanation:
a) Sample mean:
[tex]\bar{X}=\frac{1}{N} \sum X_i=\frac{1}{80}*3896.9=48.71[/tex]
Sample median: M=48.3
Note: I order the data increasingly and take the value N / 2 = 40. In this way there are 39 values above and 39 values below the median.
Sample variance:
[tex]s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41[/tex]
Sample standard deviation
[tex]s=\sqrt{s^2}=\sqrt{5.41}=2.37[/tex]
b)
The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $30,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings.
29.1 28.5 28.8 29.4 29.8 29.8 30.1 30.6
1. At the 0.05 significance level, is it reasonable to conclude the mean wedding cost is less than $30,000 as advertised?
2. State the null hypothesis and the alternate hypothesis. Use a 0.05 level of significance. (Enter your answers in thousands of dollars.)
Answer:
There is no sufficient evidence to support the claim that wedding cost is less than $30000.
Step-by-step explanation:
Values (x) ∑(Xi-X)^2
----------------------------------
29.1 0.1702
28.5 1.0252
28.8 0.5077
29.4 0.0127
29.8 0.0827
29.8 0.0827
30.1 0.3452
30.6 1.1827
----------------------------------------
236.1 3.4088
Mean = 236.1 / 8 = 29.51
[tex]S_{x}=\sqrt{3.4088/(8-1)}=0.6978[/tex]
Statement of the null hypothesis:
H0: u ≥ 30 the mean wedding cost is not less than $30,000
H1: u < 30 the mean wedding cost is less than $30,000
Test Statistic:
[tex]t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861[/tex]
Test criteria:
SIgnificance level = 0.05
Degrees of freedom = df = n - 1 = 8 - 1 = 7
Reject null hypothesis (H0) if
[tex]t<-t_{0.05,n-1}\\ t<-t_{0.05,8-1}\\ t<-t_{0.05,7}[/tex]
Finding in the t distribution table α=0.05 with df=7, we have
[tex]t_{0.05,7}=2.365[/tex]
[tex]t>-t_{0.05,7}[/tex] = -1.9861 > -2.365
Result: Fail to reject null hypothesis
Conclusion: Do no reject the null hypothesis
u ≥ 30 the mean wedding cost is not less than $30,000
There is no sufficient evidence to support the claim that wedding cost is less than $30000.
Hope this helps!
Find equation of set of points pieces that its distance from the point 3, 4, -5 and -2, 1, 4 are equal.
Answer:
Step-by-step explanation:
Suppose we a point [tex]P(x,y,z)[/tex] such that its distance from either the point [tex]A(3,4,-5)[/tex] or [tex]B(-2,1,4)[/tex] is the same.
Using this information we can formula:
distance AP = distance BP
first, let's find the distance from AP, using the distance formula.
[tex]r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}[/tex]
[tex]AP = \sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2}[/tex]
similarly, we can find the distance BP
[tex]BP = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]
since both distances are exactly the same we can equate them
[tex]AP = BP[/tex]
[tex]\sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2} = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]
we can simplify it a bit squaring both sides, and getting rid of the subscripts.
[tex](3 - x)^2 + (4 - y)^2 + (-5 - z)^2 = (-2 - x)^2 + (1 - y)^2 + (4 - z)^2[/tex]
what we have done here is formulated an equation which consists of any point P that will have the same distance from (3,4,-5) and (-2,1,4).
To put it more concretely,
This is the equation of the the plane from that consists of all points (P) from which the distance from both (3,4,-5) and (-2,1,4) are equal.
We guess, based on historical data, that 30% of graduating high-school seniors in a large city will have completed a first-year calculus course. What's the minimum sample size needed to construct a 95% confidence interval for a proportion with a margin of error of 2.5%?
Answer:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
2) Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.025[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
This exercise uses the population growth model. It is observed that a certain bacteria culture has a relative growth rate of 15% per hour, but in the presence of an antibiotic the relative growth rate is reduced to 8% per hour. The initial number of bacteria in the culture is 28. Find the projected population after 24 hours for the following conditions. (Round your answers to the nearest whole number.) (a) No antibiotic is present, so the relative growth rate is 15%. (b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
Answer:
a) P(24) = 1025.
b) P(24) = 191.
Step-by-step explanation:
This population can be modeled by the following exponential model.
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.
The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].
Population after 24 hours.
(a) No antibiotic is present, so the relative growth rate is 15%.
So r = 0.15.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]
(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
So r = 0.08.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]
Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective and why it is not injective. Hint: To show that a function f : N → N is surjective, you need to show that for all y ∈ N there is some x ∈ N such that f(x) = y. To show that a function is not injective, simply show that there are two points x1 6= x2 in the domain such that f(x1) = f(x2).
The function f(x) = x // 2 (integer division) is an example of a function from N to N that is surjective because every natural number is covered, but not injective because different numbers can result in the same output.
Explanation:To provide an example of a function f from the natural numbers N to N that is surjective but not injective, consider the function f(x) = x // 2, where '//' denotes integer division. For the function to be surjective, each element y in N must have at least one x such that f(x) = y. This is indeed the case here since for any y > 0, we can choose x = 2y or x = 2y + 1, and f(x) will equal y. To show that it is not injective, we can find two different numbers, x1 and x2, such that f(x1) = f(x2). For instance, if x1 = 4 and x2 = 5, both f(x1) and f(x2) equal 2, thus violating the definition of injectivity. Hence, f(x) = x // 2 is surjective because every y in N is an image of some x, but it is not injective because at least two different values in the domain map to the same value in the codomain.
Suppose that X has the lognormal distribution with parameters μ and σ2. Find the distribution of 1/X.
Answer:
[tex] \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]
Step-by-step explanation:
For this case we know that the distribution for the random variable is given by:
[tex]X \sim logN(\mu ,\sigma^2)[/tex]
The density function for the log normal random variable is given by:
[tex] f)x,\sigma) = \frac{1}{x \sigma \sqrt{2\pi}}e^{- \frac{ln x^2}{2\sigma^2}}[/tex]
And we want to find the distribution for the random variable [tex]\frac{1}{X}[/tex]
In order to find this distribution we can use the cumulative distribution function like this:
Let [tex] Y= \frac{1}{X}[/tex], if we solve for X from this transformation we got:
[tex] X= \frac{1}{Y}[/tex]
And then we have this:
[tex] F_Y (y) = P(Y \leq y) = P(X \leq \frac{1}{y}) = F_X (\frac{1}{y})[/tex]
And we can find the density function as the derivate of the distribution function like this:
[tex] f_Y (y) = F'_Y (y) = -\frac{1}{y^2} f_Y(\frac{1}{y})[/tex]
[tex] f_Y (y)= -\frac{1}{y^2} \frac{1}{\frac{1}{y} \sigma \sqrt{2\pi}} e^{- \frac{ln(\frac{1}{y})^2}{2\sigma^2}}[/tex]
But we see that we don't have an specified form for the distribution
If we assume that X follows a normal distribution [tex] X\sim N (\mu_z,\sigma^2_z)[/tex] and we use the transformation [tex]X=e^Y[/tex] we see that X follows a log normal distribution. And we see that:
[tex]\frac{1}{X}= \frac{1}{e^Y}=e^{-Y}[/tex]
And if we find the distribution of [tex]e^{-y}[/tex] we got this:
[tex] f_Y (y) = F'_Y (y) = -e^{-y} f_Y(e^{-y})[/tex]
[tex] f_Y (y)= -e^{-y} \frac{1}{e^{-y} \sigma \sqrt{2\pi}} e^{- \frac{ln(e^{-y})^2}{2\sigma^2}}[/tex]
[tex] f_Y (y) = -\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{y^2}{2\sigma^2}}[/tex]
And then we see that [tex]Y= \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]
help please
75
77
82
Show your work
Answer:Jarvis's class average is 75
Step-by-step explanation:
The total possible average score for the math course is 100
a) If the teacher rates homework at 10%, it means that the total possible score for homework
is 10/100 × 100 = 10
If his homework average is 93, then his score would be
(93×10)/100 = 9.3
b) If the teacher rates quizzes at 30%, it means that the total possible score for quizzes
is 30/100 × 100 = 30
If his quiz average is 82, then his score would be
(82×30)/100 = 24.6
c) If the teacher rates test at 40%, it means that the total possible score for quizzes
is 40/100 × 100 = 40
If his quiz average is 72, then his score would be
(72×40)/100 = 28.8
d) If the teacher rates final exam at 20%, it means that the total possible score for quizzes
is 20/100 × 100 = 20
If his final exam is 60, then his score would be
(60×20)/100 = 12
Jarvis's class average would be
9.3 + 24.6 + 28.8 + 12 = 74.7
Approximately 75
Average expenditure on different items such as food, clothes, fuel comes under ____. Select one: a. Descriptive statistics b. Inferential statistics c. Applied statistics d. Business statistics e. Industrial statistics
Answer:
Descriptive
Step-by-step explanation:
Statistics can be broadly classified into two main branches
i) Descriptive and ii) inferential
Descriptive statistics deal with values such as mean, standard deviation from the data.
Inferential statistics is used to predict unknown values from the observed values.
Applied statistics mainly analyses the data according to the needs of business or industries.
Hence the average expenditure on different items such as food, clothes, fuel comes under
Descriptive Statistics
Final answer:
Average expenditure on different items like food, clothes, and fuel falls under descriptive statistics, essential for summarizing trends in expenditure data. The correct option is a.
Explanation:
Descriptive statistics involve summarizing trends in data, such as calculating averages and standard deviations, making them suitable for analyzing average expenditures on items like food and clothes.
Inferential statistics are used to make inferences about populations based on sample data, aiming to find cause and effect relationships or correlations, which is essential when studying average expenditures across different categories.
Therefore, analyzing average expenditures on various items falls under the domain of descriptive statistics as it involves summarizing and interpreting trends in expenditure data.
Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.74 and plant B has a survival rate of 0.48. The standard error of the difference in proportions is 0.083. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576.
Answer:
The 99% confidence interval would be given (0.004;0.436).
In a test of hypothesis, the null hypothesis is that the population mean is equal to 74 and the alternative hypothesis is that the population mean is less than 74. A sample of 20 elements selected from this normal population produced a mean of 68.5 and a standard deviation of 6.4. The significance level is 1%. What is the value of the test statistic, t?a) 6.372
b) -4.076
c) -2.509
d) -3.843
Answer:
Step-by-step explanation:
The test statistic, t, in this hypothesis testing scenario, is calculated by using the formula t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n). Substituting given values into the formula, the answer is b) -4.076.
Explanation:In this hypothetical testing question, the test statistic, t, is calculated by dividing the difference between the sample mean and the null hypothesis mean by the standard error.
By dividing the standard deviation by the square root of the sample size, the standard error is determined. Here, the sample mean is 68.5, the null hypothesis mean is 74, the standard deviation is 6.4, and the sample size is 20.
The formula to calculate t is:
t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n)
Applying the given figures to the formula, the calculation is:
t = (68.5 - 74) / (6.4 / √20) = -4.076
So, the correct answer is b) -4.076.
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Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120 . Suppose you take a simple random sample (SRS) of 15 students from this distribution. What is the probability that a SRS of 15 students will spend an average of between 600 and 700 dollars? Round to five decimal places.
Answer:
[tex]P(600<\bar X<700)=0.89347[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the total amounts spent by students vacationing for a week in Florida of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(650,120)[/tex]
Where [tex]\mu=650[/tex] and [tex]\sigma=120[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(650,\frac{120}{\sqrt{15}})[/tex]
We are interested on this probability
[tex]P(600<\bar X<700)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(600<\bar X<700)=P(\frac{600-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{700-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(\frac{600-650}{\frac{120}{\sqrt{15}}}<Z<\frac{700-650}{\frac{120}{\sqrt{15}}})=P(-1.614<Z<1.614)[/tex]
And we can find this probability on this way:
[tex]P(-1.614<Z<1.614)=P(Z<1.614)-P(Z<-1.614)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.614<Z<1.614)=P(Z<-1.614)-P(Z<-1.614)=0.94674-0.05326=0.89347[/tex]
To find the probability, calculate the z-scores for $600 and $700, and find the area under the normal curve between these two z-scores.
Explanation:To find the probability that a simple random sample (SRS) of 15 students will spend an average of between $600 and $700, we need to calculate the z-scores for these two values.
First, we find the z-score for $600:
z = (600 - 650) / (120 / √15) = -1.5496
Next, we find the z-score for $700:
z = (700 - 650) / (120 / √15) = 1.5496
We then use a standard normal table or calculator to find the area under the normal curve between these two z-scores. The probability is the difference between these two areas.
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A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.
Answer:
The 90% confidence interval would be given by (57.006;62.994)
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X=60[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=10[/tex] represent the population standard deviation
n=30 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=10)[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=1.64[/tex]
Since we have all the values we can replace:
[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]
[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]
So on this case the 90% confidence interval would be given by (57.006;62.994)
Final answer:
To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.
Explanation:
To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:
CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)
Where:
\(\bar{x}\) is the sample mean (60 days)
\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)
\(\sigma\) is the population standard deviation (10 days)
\(n\) is the sample size (30 customers)
Plugging the values into the formula gives:
CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})
Calculating the margin of error:
ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00
Now compute the confidence interval:
CI = [60 - 3.00, 60 + 3.00]
CI = [57.00, 63.00]
So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.
Complete parts (a) through (c) below.
a) Determine the critical value(s) for a right-tailed test of a population mean at the α = 0.10 level of significance with 15 degrees of freedom.
b) Determine the critical value(s) for a left-tailed test of a population mean at the α = 0.10 level of significance based on a sample size of n = 20.
c) Determine the critical value(s) for a two-tailed test of a population mean at the α = 0.05 level of significance based on a sample size of n = 18.
Answer:
a) [tex]t_{crit}=1.34[/tex]
b) [tex]t_{crit}=-1.33[/tex]
c) [tex]t_{crit}=\pm 2.11[/tex]
Step-by-step explanation:
Part a
[tex]\alpha=0.1[/tex] represent the significance level
df =15
Since is a right tailed test the critical value is given by:
[tex]t_{crit}=1.34[/tex]
And we can use the following excel code to find it: "=T.INV(0.9,15)"
Part b
[tex]\alpha=0.1[/tex] represent the significance level
n=20 represent the sample size
First we need to find the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since is a left tailed test the critical value is given by:
[tex]t_{crit}=-1.33[/tex]
And we can use the following excel code to find it: "=T.INV(0.1,19)"
Part c
[tex]\alpha=0.05[/tex] represent the significance level
n=18 represent the sample size
First we need to find the degrees of freedom given by:
[tex]df=n-1=18-1=17[/tex]
The value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]
Since is a two tailed tailed we have two critical values is given by:
[tex]t_{crit}=\pm 2.11[/tex]
And we can use the following excel code to find it: "=T.INV(0.025,17)"
Which function represents a vertical stretch of an exponential function? f (x) = 3 (one-half) Superscript x f (x) = one-half (3) Superscript x f (x) = (3) Superscript 2 x f (x) = 3 Superscript (one-half x)
Answer:
f(x) = 3[tex](\frac{1}{2})^{x}[/tex]
Step-by-step explanation:
hope it helps!
The function that represents a vertical stretch of an exponential function is f(x) = (3)^x.
Explanation:The function that represents a vertical stretch of an exponential function is f(x) = (3)x.
In this function, the base of the exponential term is 3, and the exponent, x, determines the position on the graph. When the value of x increases, the function values also increase at an exponential rate.
For example, when x = 1, f(1) = (3)1 = 3. When x = 2, f(2) = (3)2 = 9. The function values double with each increase of x.
The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceiving a boy. As this book was being written, 51 babies were born to parents using the YSORT method, and 39 of them were boys. Use the sample data with a 0.01 significance level to test the claim that with this method, the probability of a baby being a boy is greater than 0.5. Does the method appear to work?
Answer:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]
[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]
So the p value obtained was a very low value and using 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 reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.
Step-by-step explanation:
1) Data given and notation
n=51 represent the random sample taken
X=39 represent the number of boys
[tex]\hat p=\frac{39}{51}=0.765[/tex] estimated proportion of boys
[tex]p_o=0.5[/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)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/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].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/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=0.01[/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>3.785)=7.68x10^{-5}[/tex]
So the p value obtained was a very low value and using 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 reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.
The weights of certain machine components are normally distributed with a mean of 4.81 ounces and a standard deviation of 0.04 ounces. Find the two weights that separate the top 6% and the bottom 6%. These weights could serve as limits used to identify which components should be rejected. Round your answer to the nearest hundredth, if necessary.
Suppose SAT Writing scores are normally distributed with a mean of 496 and a standard deviation of 109. A university plans to award scholarships to students whose scores are in the top 7%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.
Answer:
First question:
Top 6%: 4.87 ounces
Bottom 6%: 4.75 ounces
Second question:
Top 7%: Score of 649.4.
Step-by-step explanation:
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.
For the first problem, we have that:
[tex]\mu = 4.81, \sigma = 0.04[/tex]
Top 6%
The value of X when Z has a pvalue of 0.94. This is [tex]Z = 1.555[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.555 = \frac{X - 4.81}{0.04}[/tex]
[tex]X - 4.81 = 1.555*0.04[/tex]
[tex]X = 4.8722[/tex]
Bottom 6%
The value of X when Z has a pvalue of 0.06. This is [tex]Z = 1.555[/tex]
For the second problem, we have that:
[tex]\mu = 496, \sigma = 109[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.555 = \frac{X - 4.81}{0.04}[/tex]
[tex]X - 4.81 = -1.555*0.04[/tex]
[tex]X = 4.7477[/tex]
Top 7%
The value of X when Z has a pvalue of 0.93. This is [tex]Z = 1.475[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.475 = \frac{X - 496}{104}[/tex]
[tex]X - 496 = 104*1.475[/tex]
[tex]X = 649.4[/tex]
A disadvantage of using an arithmetic mean to summarize a set of data is that __________. Select one: a. The arithmetic mean sometimes has two values b. It can be used for interval and ratio data c. It is always different from the median d. It can be biased by one or two extremely small or large values
Answer:
d. It can be biased by one or two extremely small or large values
Step-by-step explanation:
Specially in a small sample, a single measure can cause a large difference.
For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.
So the correct answer is:
d. It can be biased by one or two extremely small or large values
The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.
Explanation:The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.
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how much mg of a metal containing 4% silver must be combined with 11 mg of a metal containing 38% silver to form an alloy containing 26% silver
Answer:
6 mg of the metal needs to be added.
Step-by-step explanation:
Let the amount (in mg) of metal that needs to be added by y.
Therefore, the amount of silver in the above metal is 0.04y.
Prior to mixing, 11 mg of a metal contained 38% of silver (Given).
Therefore, the amount of silver before= [tex]\frac{38}{100}*11[/tex]= 4.18 mg
The total amount of silver after mixing, 4.18 + 0.04y mg
The total amount of metal after mixing, 11 + y mg
New percentage of silver = 26% .
Thus, [tex]\frac{4.18+0.04y}{11+y}*100= 26[/tex]
[tex](4.18 +0.04 y) * 100 = 26 *(11 +y)[/tex]
[tex]418+4y=286+26y[/tex]
[tex]132=22y[/tex]
y=6 mg
Therefore, the amount of metal that needs to be added is 6 mg.
Suppose that for all t, a particle moving with constant speed is parameterized by r(t). Given that the length of the path from t = 5 to t = 7 is equal to 8, find the value of the speed, llv(t)ll.
Answer: value of the speed, llv(t)ll = 4.0
Step-by-step explanation:
Given;
time interval t1 =5, t2= 7
Length of path ( distance) L = 8
Speed = distance travelled/ time taken
Speed = dL/dt
Speed = 8/(t2-t1) = 8/(7-5)
Speed = 8/2
Speed = 4.0
Since it's moving at constant speed the speed = 4.0
Sheilas monthly periodic rate is 2.41%. What is her APR
Answer:
APR = 2.41% x 12 = 28.92%
Step-by-step explanation:
Her APR is 28.92%.
Sheila's APR is calculated by multiplying the monthly periodic rate of 2.41% by 12, yielding an APR of 28.92%.
Explanation:The question refers to the process of calculating an Annual Percentage Rate (APR) from a given monthly periodic rate. Sheila's monthly periodic rate is 2.41%. The APR can be calculated by multiplying this monthly rate by the number of months in a year, which is 12.
To find Sheila's APR, we perform the following calculation:
APR = Monthly Periodic Rate × Number of Periods in a Year = 2.41% × 12 = 28.92%.
Hence, Sheila's APR is 28.92%.
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A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 450 seconds and a standard deviation of 50 seconds. The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition. What time would the boys need to beat in order to earn a certificate of recognition from the fitness association? (8 pts)
Answer:
The boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 450
Standard Deviation, σ = 50
We are given that the distribution of time for fitness test is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.10
P(X<x) = 0.10
[tex]P( X < x) = P( z < \displaystyle\frac{x - 450}{50})= 0.10[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z\leq -1.282) = 0.10[/tex]
[tex]\displaystyle\frac{x - 450}{50} = -1.282\\\\x = 385.9[/tex]
Hence, boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.
The boys need to beat a time of 514 seconds, which corresponds to the 90th percentile of the normal distribution with a mean of 450 seconds and a standard deviation of 50 seconds, to be in the fastest 10% and earn a certificate of recognition.
Explanation:To determine the time the boys need to beat to be in the fastest 10% for the mile run, we need to find the corresponding z-score for the 90th percentile (since 100% - 10% = 90%) in a standard normal distribution. The z-score that correlates with the 90th percentile is approximately 1.28. We can then use the z-score formula:
Z = (X - μ) / σ
Where X is the time to beat, μ is the mean, and σ is the standard deviation. Substituting the values we know (μ = 450 seconds, σ = 50 seconds, and Z = 1.28):
1.28 = (X - 450) / 50
Multiplying both sides by 50 gives us:
64 = X - 450
Adding 450 to both sides gives:
X = 514 seconds
Therefore, boys need to beat a time of 514 seconds (or 8 minutes and 34 seconds) to earn a certificate of recognition.
Provide an appropriate response. Find the variance of the binomial distribution for which n = 800 and p = 0. 87 a. 0.90 b. 0.48 c. 696 d. 32.54 e. 9.51
Final answer:
To find the variance of a binomial distribution with n = 800 and p = 0.87, use the formula Variance = n × p × (1-p). The variance in this case is approximately 90.64.
Explanation:
To find the variance of a binomial distribution, we use the formula:
Variance = n × p × (1-p)
Where n is the number of trials and p is the probability of success.
So, in this case, with n = 800 and p = 0.87, we can calculate the variance as:
Variance = 800 × 0.87 × (1-0.87)
Variance = 800 × 0.87 × 0.13
Variance = 90.64
Therefore, the variance of the binomial distribution is approximately 90.64.
A few years ago, a census bureau reported that 67.4% of American families owned their homes. Census data reveal that the ownership rate in one small city is much lower. The city council is debating a plan to offer tax breaks to first-time home buyers in order to encourage people to become homeowners. They decide to adopt the plan on a 2-year trial basis and use the data they collect to make a decision about continuing the tax breaks. Since this plan costs the city tax revenues, they will continue to use it only if there is strong evidence that the rate of home ownership is increasing.
Who would be harmed by a Type II error?
(A) The city, because it would lose tax revenue. Faster pace
(B) The citizens of the city, because they lose help they could have used to buy a home.
(C) The city, because it would lose homeowners.
(D) The citizens of the city, because they would have to pay higher taxes than before.
(E) There is no Type Il error in this context.
Answer:
(B) The citizens of the city, because they lose help they could have used to buy a home.
Step-by-step explanation:
Nul and alternative hypotheses are:
[tex]H_{0}:[/tex] the rate of home ownership is the same after tax cut[tex]H_{a}:[/tex] the rate of home ownership is increasing after tax cutType II error occurs when one fails to reject null hypothesis when the null hypothesis was wrong.
In this case Type II error happens when the conclusion is the rate of home ownership is not increasing after tax cut, where actually it is.
With this conclusion city council does not continue tax cut, and citizens of the city is harmed because they lose help they could have used to buy a home.
Final answer:
The citizens of the city would be harmed by a Type II error as they would miss out on the help to buy homes.
Explanation:
Who would be harmed by a Type II error?
(B) The citizens of the city, because they lose help they could have used to buy a home.
A Type II error in this context would harm the citizens of the city as they would miss out on the intended help in buying homes due to the failure to detect an increase in the rate of home ownership.
Find the sample size, n, needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.02 margin of error, use a confidence level of 98% and use results from a prior poll suggesting that 20% of adults have consulted fortune tellers.
Answer: 2172
Step-by-step explanation:
Formula to find the sample size n , if the prior estimate of the population proportion(p) is known:
[tex]n= p(1-p)(\dfrac{z^*}{E})^2[/tex] , where E= margin of error and z* = Critical z-value.
Let p be the population proportion of adults have consulted fortune tellers.
As per given , we have
p= 0.20
E= 0.02
From z-table , the z-value corresponding to 98% confidence interval = z*=2.33
Then, the required sample size will be :
[tex]n= 0.20(1-0.20)(\dfrac{2.33}{0.02})^2[/tex]
[tex]n= 0.20(0.80)(116.5)^2[/tex]
[tex]n= 2171.56\approx2172[/tex]
Hence, the required sample size = 2172
A continuous random variable may assume :
a. any numerical value in an interval or collection of intervals.b. finite number of values in a collection of intervals.c. an infinite sequence of values.d. only the positive integer values in an interval.
Answer:
Option A) any numerical value in an interval or collection of intervals
Step-by-step explanation:
Continuous Random Variable:
A continuous random variable can take any value within an interval.Thus, it can take infinite values since there are infinite numbers in an interval.A continuous variable is a variable whose value is obtained by measuring.Examples: height of students in class , weight of students in class, time it takes to get to school, distance traveled between classes.Thus, the correct meaning of continuous random variable is explained by Option A)Option A) any numerical value in an interval or collection of intervals
A continuous random variable can take on any numerical value within a given range or collection of ranges, and it's a characteristic feature of it to take an infinite number of values in any interval. Some examples of this can be a person's height, time, temperature, and weight in physics.
Explanation:A continuous random variable is a type of random variable that can assume any numerical value in a given interval or collection of intervals, making option a correct. This is in contrast to a discrete random variable, which can only take on a finite number of values. A key characteristic of continuous random variables is that they can take on an infinite number of values in any interval.
For example, the height of a person can be treated as a continuous variable, since it can theoretically take any value within a certain range, not just whole number values. The same applies to variables such as time, temperature, and weight in physics.
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