The correct conclusion is to not reject the null hypothesis.
Explanation:If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to not reject the null hypothesis.
When conducting a hypothesis test, the p-value represents the probability of obtaining the observed data or more extreme values assuming the null hypothesis is true. If the p-value is greater than the significance level (α), which is the threshold for rejecting the null hypothesis, we fail to reject the null hypothesis.
In this case, the p-value of 0.07 is greater than the significance level of 0.05. Therefore, we cannot reject the null hypothesis.
A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the population was 158.
1. What was the population of the village at the beginning of the 10 years according to the exponential growth function? Round your answer up to the next whole number, and do not include units.
Answer:
The initial population at the beginning of the 10 years was 129.
Step-by-step explanation:
The population of the village may be modeled by the following function.
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, in decimal.
In this problem, we have that:
[tex]P(10) = 158, r = 0.02[/tex].
So
[tex]158 = P_{0}e^{0.02*10}[/tex]
[tex]P_{0} = 158*e^{-0.2}[/tex]
[tex]P_{0} = 129[/tex]
The initial population at the beginning of the 10 years was 129.
The population of the village at the beginning of the 10 years was approximately 130. This is determined by using the formula for exponential growth that is continuously compounded and solving for the initial population.
Explanation:We can use the formula for exponential growth that is continuously compounded, which is P=Pe^rt, where P is the final population, P is the initial population, r is the growth rate, and t is the time.
In this case, we know that P=158, r=0.02 (2% expressed as a decimal), and t=10. We need to find P.
Substituting the given values into the formula, we get 158=Pe^(0.02*10).
We can calculate e^(0.02*10) using a calculator to get approximately 1.22. Therefore, the equation becomes 158=P*1.22.
To find P, we can divide both sides of the equation by 1.22 resulting in P ≈ 158/1.22 ≈ 129.5. Since we are asked to round up to the next whole number, the original population was approximately 130.
Learn more about Exponential GrowthA coin is tossed 3 times. What is the probability that the number of tails obtained will be between 1 and 3 inclusive?
Answer: [tex]\dfrac{7}{8}[/tex]
Step-by-step explanation:
Given : When a coin is tossed twice , the total outcomes will become
HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT
Total outcomes = 8
Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT
Number of favorable outcomes = 7
Probability that the number of tails obtained will be between 1 and 3 inclusive = [tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]=\dfrac{7}{8}[/tex]
Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive [tex]=\dfrac{7}{8}[/tex]
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]
Question 5. We have 5-year statistics of average amount of wheat crop (tons) harvested from 1 km2 per year, the results are as follows: 560, 525, 496, 543, 499. Test the hypothesis that the mean wheat crop is 550 tons per 1km2 (α = 0.05) and choose the correct answer. Determine a 95% confidence interval on the mean wheat crop. Determine whether the hypothesis that the mean crop is 550 tons p
Answer:
a) [tex]t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]
[tex]p_v =P(t_{(4)}<-2.052)=0.055[/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.
We can say that at 5% of significance the true mean is not significantly different from 550.
b) The 95% confidence interval would be given (490.184;559.016).
We are confident at 95% that the true mean is between (490.184;559.016).
Step-by-step explanation:
Part a
Data given and notation
560, 525, 496, 543, 499
The mean and sample deviation can be calculated from the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}[/tex]
[tex]\bar X=524.6[/tex] represent the sample mean
[tex]s=27.682[/tex] represent the sample standard deviation
[tex]n=5[/tex] sample size
[tex]\mu_o =550[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t 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 population mean is equal to 550, the system of hypothesis are:
Null hypothesis:[tex]\mu = 550[/tex]
Alternative hypothesis:[tex]\mu \neq 550[/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{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]
P-value
We need to calculate the degrees of freedom first given by:
[tex]df=n-1=5-1=4[/tex]
Since is a one-side left tailed test the p value would given by:
[tex]p_v =P(t_{(4)}<-2.052)=0.055[/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.
We can say that at 5% of significance the true mean is not significantly different from 550.
Part b
The confidence interval for a proportion is given by this formula
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
The degrees of freedom are 4 from part a
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]t_{\alpha/2}=2.78[/tex]
And replacing into the confidence interval formula we got:
[tex]524.6 - 2.78 \frac{27.682}{\sqrt{5}}=490.184[/tex]
[tex]524.6 + 2.78 \frac{27.682}{\sqrt{5}}=559.016[/tex]
And the 95% confidence interval would be given (490.184;559.016).
We are confident at 95% that the true mean is between (490.184;559.016).
Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.
Before 10 12 14 16 12
After 15 14 17 17 20
The obtained value of the appropriate statistic is ____.
a. 3.92
b. 2.58
c. 4.12
d. 3.06
What do you conclude using a = 0.052 tail?
a. retain H0; we cannot conclude that the class improved study habits
b. accept H0; the class appeared to improve study habits
c. retain H0; the class had no effect on study habits
d. reject H0; the class appeared to improve study habits
Answer:
The obtained value of the appropriate statistic is ____.
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]
d. 3.06
[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]
d. reject H0; the class appeared to improve study habits
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.
Let put some notation
x=before method , y = after method
x: 10 12 14 16 12
y: 15 14 17 17 20
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]
The first step is define the difference [tex]d_i=y_i-x_i[/tex], that is given so we have:
d: 5,2,3,1,8
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=3.8[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =2.775[/tex]
The fourth step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=5-1=4[/tex]
Now we can calculate the p value, since we have a two tailed test the p value is given by:
[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]
The p value is less than the significance level given [tex]\alpha=0.05[/tex], so then we can conclude that we reject the null hypothesis.
d. reject H0; the class appeared to improve study habits
You are given the polar curve r = e^θ
List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of r and limit yourself to 1<_r<_1000 ( note the restriction on r) and 0≤θ<2????. If two or more points share the same value of r, list those starting with the smallest value of θ. If any blanks are unused, type an upper-case "N" in them.
Point 1: (r, θ) = ________
Point 2: (r, θ) = ________
Point 3: (r, θ) = ________
The tangent to [tex]r(\theta)=e^\theta[/tex] has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], where
[tex]\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}[/tex]
By the chain rule, we have
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}[/tex]
and by the product rule,
[tex]\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta[/tex]
so that with [tex]\frac{\mathrm dr}{\mathrm d\theta}=e^\theta[/tex], we get
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}[/tex]
The tangent line is horizontal when the slope is 0; this happens for
[tex]-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi[/tex]
where [tex]n[/tex] is any integer. In the interval [tex]0\le\theta\le2\pi[/tex], this happens for [tex]n=1,2[/tex], or
[tex]\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4[/tex]
i.e at the points
[tex](r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)[/tex]
and
[tex](r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)[/tex]
Direct mail advertisers send solicitations ("junk mail") to thousands of potential customers in the hope that some will buy the company's product. The response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1140 people randomly selected from their mailing list of over 200,000 people. They get orders from 122 of the recipients. Use this information to complete parts a through d.Create a 95% confidence interval for the percentage of people the company contacts who may buy something. (Show your work. Step by step)
Answer: (0.089, 0.125)
Step-by-step explanation:
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size.
[tex]\hat{p}[/tex] = Sample proportion.
z*= Critical z-value.
Let p be the population proportion of people the company contacts who may buy something.
As per given , sample size : n= 1140
Number of recipients ordered = 122
Then, [tex]\hat{p}=\dfrac{122}{1140}\approx0.107[/tex]
Critical value for 95% confidence interval = z*= 1.96 (By z-table)
So , the 95% confidence interval for the percentage of people the company contacts who may buy something:
[tex]0.107\pm (1.96)\sqrt{\dfrac{0.107(1-0.107)}{1140}}[/tex]
[tex]=0.107\pm (1.96)\sqrt{0.000083817}[/tex]
[tex]=0.107\pm (1.96)(0.00915516)[/tex]
[tex]=0.107\pm 0.018[/tex]
[tex]=(0.107-0.018,\ 0.107+0.018)=(0.089,\ 0.125)[/tex]
Hence, the 95% confidence interval for the percentage of people the company contacts who may buy something = (0.089, 0.125)
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags.
A 26 bag sample had a mean of 427 grams with a standard deviation of 15.
Assume the population is normally distributed. A level of significance of 0.05 will be used.
Find the P-value of the test statistic.
You may write the P -value as a range using interval notation, or as a decimal value rounded to four decimal places.
Answer:
P-value = 0.0261
We conclude that the machine is under-filling the bags.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 433 gram
Sample mean, [tex]\bar{x}[/tex] = 427 grams
Sample size, n = 26
Alpha, α = 0.05
Sample standard deviation, σ = 15 grams
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 433\text{ grams}\\H_A: \mu < 433\text{ grams}[/tex]
We use one-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{427 - 433}{\frac{15}{\sqrt{26}} } = -2.0396[/tex]
Now, we calculate the p-value using the standard table.
P-value = 0.0261
Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.
We conclude that the machine is under-filling the bags.
more thanAn engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 120 engines and the mean pressure was 5 lbs/square inch. Assume the variance is known to be 1. If the valve was designed to produce a mean pressure of 5.1 lbs/square inch, is there sufficient evidence at the 0.02 level that the valve performs below the specifications? State the null and alternative hypotheses for the above scenario. Answer
Answer:
We accept the null hypothesis
Step-by-step explanation:
Given
The valve was tested on 120 engines
Mean pressure = 5 lbs/square inch.
Variance = 1
Standard Deviation, σ = √1 = 1
The valve was designed to produce a mean pressure of 5.1 lbs/square inch
So, μ = 5.1
Null hypothesis: H₀ : μ = 5.1
Alternative Hypothesis: H₁ : μ≠ 5.1
Since n > 30 and population standard deviation is given
So, We will use z test
Formula : z = (x - μ)/( σ /√n) ---_-_--- Substitute the values
z = [tex]\frac{5 - 5.1}{1/\sqrt{120} }[/tex]
z = −1.1
The p - value of -1.1 (refer to z table) is 0.13576
Since it is a two tailed test So, p = 2(1- 0.13576) = 1.7285
α = 0.02
p value > α
So, we accept the null hypothesis
Hence There is no sufficient evidence at the 0.02 level that the valve does not performs below to the specifications
A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.
The group of 51 students in the study reported an average of 4.35 drinks per with a standard deviation of 3.88 drinks.
Find the p-value for the hypothesis test.
The p-value should be rounded to 4-decimal places.
Answer:
P-value = 0.4846
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4.73
Sample mean, [tex]\bar{x}[/tex] = 4.35
Sample size, n = 51
Alpha, α = 0.05
Sample standard deviation, s = 3.88
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73[/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{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]
We can calculate the p-value with the help of standard normal table.
P-value = 0.4846
Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.
We conclude that this college has same drinking habit as the college students in general.
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).
A probability distribution showing the probability of success does not change from trial to trial, is termed a:
-uniform probability distribution
-binomial probability distribution
-hypergeometric probability distribution
-normal probability distribution
Answer:
-uniform probability distribution
Step-by-step explanation:
In uniform probability distributions, the likelihood of each possible outcome happening or not is the same. This property means that, for any given trial, the probability that an event will be successful does not change. Take the probability of rolling a 5 on a die for instance, no matter how many trials are performed, there is always a 1 in 6 probability for each trial.
A probability distribution showing the probability of success does not change from trial to trial, is termed as a uniform probability distribution.
We have to determine, a probability distribution showing the probability of success does not change from trial to trial, is termed as;
A continuous probability distribution is called the uniform distribution and it is related to the events that are equally possible to occur. It is defined by two different parameters, x, and y,
Where x = the minimum value and y = the maximum value. It is generally represented by u(x, y).
The case of a discrete binomial probability distribution. The Bernoulli trials are identical but independent of each other. Before computing the failures (“r”), the total count of success that occurs first is called the Negative Binomial Probability Distribution.
Probability Distributions give up the possible outcome of any random event. It is also identified on the grounds of underlying sample space as a set of possible outcomes of any random experiment.
The normal distribution is a probability distribution that outlines how the values of a variable are distributed. Most of the observations cluster around the central peak and probabilities for values far away from the mean fall off equally in both directions in a normal distribution. Extreme values in both the tails of the distribution are uniformly unexpected.
Hence, A probability distribution showing the probability of success does not change from trial to trial, is termed as a uniform probability distribution.
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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!
The Pew Research Center Internet Project conducted a survey of 957 Internet users. This survey provided a variety of statistics on them. If required, round your answers to four decimal places. (a) The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally.
Answer: (0.881, 0.919)
Step-by-step explanation:
Let p be the population proportion of respondents who say the Internet has been a good thing for them personally.
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where z* = Critical value ,
[tex]\hat{p}[/tex] = Sample proportion
n = sample size.
As per given , we have
n= 957
[tex]\hat{p}=0.90[/tex]
By z-table , Critical value for 95% confidence interval : z*= 1.96
Then, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. will be :
[tex]0.90\pm 1.96\sqrt{\dfrac{0.90(1-0.90)}{957}}[/tex]
[tex]0.90\pm 1.96\sqrt{0.0000940438871473}[/tex]
[tex]0.90\pm 1.96(0.00969762275753)[/tex]
[tex]\approx0.90\pm 0.019\\\\=(0.90-0.019,\ 0.90+0.019)[/tex]
[tex](0.881,\ 0.919)[/tex]
Hence, the required 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally = (0.881, 0.919)
To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula: CI = p ± Z * sqrt((p * (1 - p)) / n). Plugging in the values, we can calculate a 95% confidence interval of [0.8808, 0.9192].
Explanation:To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula:
CI = p ± Z * sqrt((p * (1 - p)) / n)
Where:
CI is the confidence intervalp is the proportion of respondents who say the Internet has been a good thing for them (0.90)Z is the z-score corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96)n is the sample size (957)Plugging in the values, we can calculate:
CI = 0.90 ± 1.96 * sqrt((0.90 * (1 - 0.90)) / 957)
This gives us a 95% confidence interval for the proportion, which is [0.8808, 0.9192]. So we can be 95% confident that the true proportion of respondents who say the Internet has been a good thing for them is between 0.8808 and 0.9192.
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The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. The manufacturer's specifications state that the standard deviation of the amount of paint is equal to 0.02 gallon. A random sample of 50 cans is selected, and the sample mean amount of paint per 1-gallon can is 0.998 gallon. Complete parts (a) through (d).
a. Construct a 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can.
b. On the basis of these results, do you think the manager has a right to complain to the manufacturer? Why?
___because a 1-gallon paint can containing exactly 1-gallon of paint lies _____the 95% confidence interval.
c. Must you assume that the population amount of paint per can is normally distributed here? Explain.
d. Construct a 90% confidence interval estimate. How does this change your answer to part (b)?
How does this change your answer to part (b)?
A 1-gallon paint can containing exactly 1-gallon of paint lies ____the 90% confidence interval. The manager _______a right to complain to the manufacturer.
Answer:
a. 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can is 0.998±0.0055
b. No, because a 1-gallon paint can containing exactly 1-gallon of paint lies within the 95% confidence interval.
c. Yes. The population amount of paint per can is assumed normally distributed, because confidence interval calculations assume normal distribution of the parameter.
d. 90% confidence interval is 0.998±0.0046. The answer in b. didn't change; 1-gallon paint can containing exactly 1-gallon of paint lies within the 90% confidence interval. The manager doesn't have a right to complain to the manufacturer.
Step-by-step explanation:
Confidence Interval can be calculated using M±ME where
M is the sample mean amount of paint per 1-gallon can (0.998 gallon)
ME is the margin of error from the mean
And margin of error (ME) can be calculated using the equation
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in the 95% confidence level (1.96) s is the sample standard deviation (0.02 gallon)N is the sample size (50)Then ME=[tex]\frac{1.96*0.02}{\sqrt{50} }[/tex]≈0.0055
95% confidence interval is 0.998±0.0055
90% confidence interval can be calculated similary, only z statistic is 1.64.
ME=[tex]\frac{1.64*0.02}{\sqrt{50} }[/tex] ≈0.0046
90% confidence interval is 0.998±0.0046
Use Cramer's Rule to find x in the system of equations below.
2x − 3y = 17
5x + 4y = 8
Answer: x = 4 , y = -3
Step-by-step explanation:
Going by the Cramer's rule , we first determine the determinant by dealing with only the coefficients of x and y in the 2 x 2 matrix.
2x - 3y = 17
5x + 4y = 8
2 -3
5 4, going by the rule now, we now have
(2 x 4) - (5 x -3)
8 + 15
= 23.
Now to find the value of x , replace the constants with the coefficient of x and divide by he determinants.
17 -3
8 4
---------------
2 -3
5 4
( 17 x 4 ) - ( 8 x -3 )
---------------------------
23
= 68 + 24
------------
23
= 92/23
= 4.
x = 4
Now to find y, just repeat the process by replacing the coefficient of y with the constants.
2 17
5 8
-----------
2 -3
5 4
( 2 x 8 ) - ( 5 x 17 )
-----------------------
23
16 - 85
---------
23
= -69/23
= -3
y = -3.
check
substitute for the values in any of the equations above.
2(4) - 3(-3)
8 + 9
= 17
Biologist Theodore Garland, Jr. studied the relationship between running speeds and morphology of 49 species of cursorial mammals (mammals adapted to or specialized for running). One of the relationships he investigated was maximal sprint speed in kilometers per hour and the ratio of metatarsal-to-femur length. A least-squares regression on the data he collected produces the equation ^ y = 37.67 + 33.18 x where x is metatarsal-to-femur ratio and ^ y is predicted maximal sprint speed in kilometers per hour. The standard error of the intercept is 5.69 and the standard error of the slope is 7.94. Construct a 96% confidence interval for the slope of the population regression line. Give your answers precise to at least two decimal places.
Answer:
Confidence interval for the intercept:
[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]
Confidence interval for the slope:
[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]
Step-by-step explanation:
We start defining our equation's terms, starting from the linear regression model [tex]\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x[/tex]
In this model [tex]{\beta}_{0} [/tex] is the intercept estimator and [tex]{\beta}_{1} [/tex] is the slope estimator.
in the equation y = 37.67 + 33.18x]
[tex]{\beta}_{0} = 37.67 [/tex] and [tex]{\beta}_{1} = 33.81[/tex]
Then we have the standard errors (se) for each estimator:
[tex]se(\hat{\beta_{0}}}) = 5.69[/tex] and [tex]se(\hat{\beta_{1}}}) = 7.94[/tex]
The sample number is 49 species (here we assume that all the individuals of a the same species are summarised with a central tendency measure, i.e. mean, median or mode, if each species contained more than one individual).
The formula for the 96% confidence interval for the intercept [tex]{\beta}_{0} [/tex] we have:
[tex]\hat{\beta_0} - t_{\alpha/2,n-2} se(\beta_0) \leq \beta _0 \leq \hat{\beta_0} + t_{\alpha/2,n-2}[/tex]
Where [tex]t_{\alpha/2,n-2}[/tex] represents the p-value in a t distribution when α=0.04 (so that we have a confidence interval of 96%, or 0.96), two-tailed, and n-2 degrees of freedom. In this example, n-2 = 47, and the t-value (47 degrees of freedom, 0.04 significance level, two tails) is ± 2.1123.
We input these values into our formula:
[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]
Similarly, the 96% confidence interval for the slope [tex]{\beta}_{1} [/tex] is:
[tex]\hat{\beta_1} - t_{\alpha/2,n-2} se(\beta_1) \leq \beta _1 \leq \hat{\beta_1} + t_{\alpha/2,n-2}[/tex]
Where [tex]t_{\alpha/2,n-2} =± 2.1123[/tex]
And into the formula:
[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]
The confidence interval does not include 0, so there is enough evidence saying that there is enough correlation between the metatarsal-to-femur ratio and maximal sprint speed in kilometers per hour. This study shows that measuring the lengths of metatarsal 3 and femur in mammals is a reliable predictor of maximal speed for cursorial mammals.
The 96% confidence interval for the slope of the population regression line is calculated using the formula b ± t(α/2, df) * SE(b). Plugging in the given values (slope= 33.18, standard error= 7.94) along with t(0.02, 47) which is approximately ± 2.01, we get the interval (16.98, 49.38).
Explanation:In statistics, the confidence interval for the slope of the population regression line can be estimated by using the following formula: b ± t(α/2, df) * SE(b), where b is the slope, t is the t-score, df represents the degrees of freedom, α is the significance level, and SE(b) is the standard error of the slope.
In this case, we are asked to construct a 96% confidence interval for the slope, so α = 1 - 0.96 = 0.04. Since we are dealing with a two-tailed test, we will distribute α across the two tails, giving us α/2 = 0.02 for each half. To calculate our t-score, we will use the t distribution table and look for the value that corresponds with df = n - 2 (since we have two parameters, the intercept and slope) and α/2 = 0.02. Given we have 49 species, the df = 49 - 2 = 47, the t-score is approximately ± 2.01.
With this information, we can substitute the values into the formula, where b = 33.18 and SE(b) = 7.94, and compute the confidence interval for the slope, giving us 33.18 ± 2.01 * 7.94 = (16.98, 49.38). Thus, we estimate with 96% confidence that the true population slope lies between 16.98 and 49.38.
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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.
Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.2. A university plans to offer tutoring jobs to students whose scores are in the top 1%.What is the minimum score required for the job offer? Round your answer to 1 decimal place.
Answer:
So the value of score that separates the bottom 99% of data from the top 1% is 33.316.
And rounded to 1 decimal place would be 33.3
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".
2) Solution to the problem
Let X the random variable that represent the scores for the ACT of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(21.5,5.2)[/tex]
Where [tex]\mu=21.5[/tex] and [tex]\sigma=5.2[/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}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.01[/tex] (a)
[tex]P(X<a)=0.99[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(Z>2.33)=0.01
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99[/tex]
[tex]P(Z<\frac{a-\mu}{\sigma})=0.99[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]Z=2.33<\frac{a-21.2}{5.2}[/tex]
And if we solve for a we got
[tex]a=21.2 +2.33*5.2=33.316[/tex]
So the value of score that separates the bottom 99% of data from the top 1% is 33.316.
And rounded to 1 decimal place would be 33.3
According to the Mendelian theory of genetics, a certain garden pea plant should produce either white, pink, or red flowers with respective probabilities 1/4, 1/2, and 1/4. To test this theory, a sample of 564 peas was conducted with the result that 141 produced white, 291 produced pink, and 132 produced red flowers. Using the chi-square approximation, what conclusion would be drawn at the 5 percent level of significance?
Answer:
[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]
[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(0.861,2,TRUE)"
Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Solution to the problem
Assume the following dataset:
White = 141, Pink = 291, Red= 132
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]
H1: There is difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]
The level of significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = p_i *Total[/tex]
And the calculations are given by:
[tex]E_{White} =\fra{1}{4} * 564=141[/tex]
[tex]E_{Pink} =\frac{1}{2} *564=282[/tex]
[tex]E_{Red} =\frac{1}{4}*564=141[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=categories-1=3-1=2[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(0.861,2,TRUE)"
Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.
A small entrepreneurial company is trying to decide between developing two different products that they believe they can sell to two potential companies, one large and one small. If they develop Product A, they have a 50% chance of selling it to the large company with annual purchases of about 20,000 units. If the large company won't purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units. On the other hand if they develop Product B, they feel they have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn't buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units.
1-What is the probability that Product B will being purchased by the smaller company?
A. 0.5
B. 0.4
C. 0.3
D. 0.8
Answer:
The correct option is C) 0.3
Step-by-step explanation:
Consider the provided information.
A small entrepreneurial can develop two product, product A and product B.
If they develop Product A:
There are 50% chance of selling it to the large company with annual purchases of about 20,000 units.
If the large company won't purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units.
If they develop Product B.
They have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn't buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units.
If they develop product B than the probability of purchasing by small company is 60% or 0.6
If large company doesn't buy it, then they have 50%or 0.5 chance of selling it to small company.
Hence, the total probability is: [tex]0.6\times0.5=0.3[/tex]
Therefore, the correct option is C) 0.3
Final answer:
The probability that Product B will be purchased by the smaller company is 30%, which corresponds to answer choice C.
Explanation:
The probability that Product B will be purchased by the smaller company is calculated by considering that Product B will only be offered to the smaller company if the large company does not purchase it. So, first, we must calculate the probability that the large company does not purchase Product B, which is 1 - 0.4 (since there is a 40% chance that the large company will purchase it). This results in a 60% chance that the large company will not purchase Product B. The probability that Product B will then be purchased by the smaller company is 50% of that 60%, which is calculated as 0.6 × 0.5 = 0.3 or 30%.
Show that the class of closed intervals [a, b], where a is rational and b is irrational and a < b, is a base for a topology on the real line R.
Answer:
See proof below.
Step-by-step explanation:
Remember that a collection of subsets B of a set X is a base for a topology of X if the following conditions hold
The union of B is equal to XIf T,S∈B and p∈T∩S then there exists some Q∈B such that x∈Q and Q⊆T∩S.In this case, X=R and B={[a,b]: a is rational and b is irrational}. Let's prove the statements above
The inclusion ∪B⊆R is true for any collection of subsets of the real line. To prove that R⊆∪B., let x∈R. If x is rational choose an irrational number y>x. (this can be done because irrational numbers are not bounded above) Then x∈[x,y] and [x,y]∈B, thus x∈∪B. If x is irrational, choose some rational number z<x (it can be done because rationals are not bounded below). Then x∈[z,x] for some [z,x]∈B, thus by definition of union, x∈∪B. In any case, x∈∪B therefore R=∪B.Let T=[a,b], S=[c,d]∈B be arbitrary elements of B. Suppose that p∈T∩S. Define Q=T∩S. Q is a closed interval, its starting point is either a or c (the greatest of these), which are rational, and its endpoint is either b or d (the smallest of these), which are irrational. In any case Q∈B and we have that p∈Q⊆T∩S.which of the following statement is not true?
integers are closed under division
whole numbers are not closed under subtraction
whole numbers are closed under addition
natural numbers are closed under multiplication
Answer:
integers are closed under division
Step-by-step explanation:
Integers are not closed under division.Let us consider two integers i and j.
Then [tex]\[\frac{i}{j}\][/tex] need not necessarily be an integer. For example, 2 and 3 are integers but [tex]\[\frac{2}{3}\][/tex] is not an integer. So the first option , namely, integers are closed under division is not true.
On the other hand, the remaining three options given are correct.
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
You’re at a party with 199 other guests when robbers break in and announce that they are going to rob one of you. They put 199 blank pieces of paper in a hat, plus one marked "you lose." Each guest must draw, and the person who draws "you lose" will get robbed. The robbers o↵er you the option of drawing first, last, or at any time in between. When would you take your turn?
Answer:
First
Step-by-step explanation:
Since, the probability that the first person pick the paper marked "you lose" is 1/200, which is smaller than the probability of who draws later.
Taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.
The optimal strategy in this scenario is to draw last.
Here's the reason:
When drawing first, you have a 1/200 chance of drawing the "you lose" paper initially.
However, if you draw last, you get to observe the outcomes of all the previous draws.
If none of the previous guests has drawn the "you lose" paper, then the odds of it being in the hat when it's your turn are 1/200.
In contrast, if someone before you draws the "you lose" paper, you won't have to draw at all since the game ends. This means that by drawing last, you have the best chance of avoiding the "you lose" paper if others before you didn't draw it.
Thus, taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.
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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
This probability distribution shows the typical grade distribution fr a Geometry course with 35 students.
Using the probabilities given, find the probability that a student earns a grade of A.
p=
Enter a decimal rounded to the nearest hundredth.
Answer:
The probability that a student earns a grade of A is 1/7.
Step-by-step explanation:
Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:
P(E) = n(E) / n(S)
As the total number of students = n(S) = 35
Students getting the grade A = n(E) = 5
So, the probability that a student earns a grade of A:
P(E) = n(E) / n(S)
= 5/35
= 1/7
Hence, the probability that a student earns a grade of A is 1/7.
Keywords: probability, sample space, event
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Answer:
.14
Step-by-step explanation:
A sample of 106 healthy adults have a mean body temperature of 98.2, and a standard deviation of 0.62. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. Group of answer choices ( 98.04, 98.36 ) ( 97.54, 97.95 ) ( 97.84, 98.12 ) ( 97.82, 98. 15 ) ( 97.95, 98.15 )
Answer: [tex](98.04,\ 98.36)[/tex]
Step-by-step explanation:
Given : Sample size of healthy adults: n= 106
Degree of freedom = df =n-1 = 105
Sample mean body temperature : [tex]\overline{x}=98.2[/tex]
Sample standard deviation : [tex]s= 0.62[/tex]
Significance level ; [tex]\alpha= 1-0.99=0.01[/tex]
∵ population standard deviation is unknown , so we use t- critical value.
Confidence interval for the population mean :
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value for df = 105 and [tex]\alpha=0.01[/tex]= [tex]t_{\alpha/2, df}=t_{0.005 , 105}=2.623[/tex]
A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :
[tex]98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}[/tex]
[tex]98.2\pm (2.623)(0.0602197234662)[/tex]
[tex]98.2\pm (0.157956334652)[/tex]
[tex]\approx98.2\pm 0.16[/tex]
[tex](98.2-0.16,\ 98.2+0.16)[/tex]
[tex](98.04,\ 98.36)[/tex]
Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans = [tex](98.04,\ 98.36)[/tex]
What is this describing: The set of all points between R and S including R and S.
Line Segment RS
Line RS
Ray SR
Ray RS
Need answer rn !!!!
Answer:
Line segment RS
Step-by-step explanation:
Line segment RS includes all points between R and S including R and S.Line RS contains all points between R and S (including R and S) and also the points in the extended line segment RS both sides to infinity.Ray SR starts from S and extends to infinity through R.Ray RS starts from R and extends to infinity through S.Hence the correct answer is Line segment RS.
In a Harris poll, adults were asked if they are in favor of abolishing the penny. Among the responses, 1241 answered "no," 459 answered "yes," and 352 had no opinion. What is the sample proportion of yes responses, and what notation is used to represent it? A.ModifyingAbove p with caret p=0.270. The symbol ModifyingAbove p with caret p is used to represent a sample proportion. B.p=0.224. The symbol p is used to represent a sample proportion. C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion. D.p=0.270. The symbol p is used to represent a sample proportion.
Answer:
C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.
Step-by-step explanation:
Data given and notation
X= 459 represent the number of people that answer yes
1241 people answer no and 352 had no opinion
n= 459+1241+352=2052 represent the sampplpe size selected
Solution to the problem
What is the sample proportion of yes responses, and what notation is used to represent it?
If we want to find the sample proportion of yes we need to apply the following formula:
[tex]\hat p =\frac{X}{n}[/tex]
Where X represent the number of people with the characteristic desired in the sample and n the sample size selected.
For this case we have:
[tex]\hat p =\frac{X}{n}=\frac{459}{2052}=0.224[/tex]
Let's analyze the possible options given:
A.ModifyingAbove p with caret p=0.270. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.
False the sample proportion is 0.224 not 0.270
B.p=0.224. The symbol p is used to represent a sample proportion.
The sample proportion is 0.224 not the population proportion for this case this statement is False.
C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.
Correct the [tex]\hat p= 0.224[/tex] represent th sample proportion and it's the correct way to express it.
D.p=0.270. The symbol p is used to represent a sample proportion.
False the sample proportion is 0.224 not 0.270
To find the sample proportion of yes responses, divide the number of yes responses by the total responses. The sample proportion is 0.224, represented as p'. The correct answer is option C.
To determine the sample proportion of yes responses in the Harris poll, we need to use the formula for sample proportion:
p' = x/n, where x represents the number of successes (yes responses) and n represents the sample size (total responses).
Step-by-Step:
Total number of responses = 1241 (no) + 459 (yes) + 352 (no opinion) = 2052Number of yes responses (x) = 459Sample proportion (p') = 459 / 2052 ≈ 0.224The sample proportion of yes responses is 0.224. In statistical notation, this is represented as p'.
The correct answer is: C. p' = 0.224. The symbol p' is used to represent a sample proportion.