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Exam 9: Inferences From Two Samples

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Question 91

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Which distribution is used to test the claim that the standard deviation of the ages (in years) of when girls first learn to ride a bike is equal to the standard deviation of the ages (in years) when boys first lean to ride a bike?

A) F
B) chi-square
C) t
D) Normal

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Question 92

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Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. -The manager of a juice bottling factory is considering installing a new juice bottling machine which she hopes will reduce the amount of variation in the volumes of juice dispensed into 8-fluid-ounce bottles. Random samples of 10 bottles filled by the old machine and 9 bottles filled by the new machine yielded the following volumes of juice (in fluid ounces). Old machine: $\quad 8.2,8.0,7.9,7.9,8.5,7.9,8.1,8.1,8.2,7.9$ New machine: $8.0,8.1,8.0,8.1,7.9,8.0,7.9,8.0,8.1$ Use a 0.05 significance level to test the claim that the volumes of juice filled by the old machine vary more than the volumes of juice filled by the new machine. $\left( \text { Note } : \mathrm { s } _ { 1 } = 0.1947 \mathrm { fl } \mathrm { oz } , \mathrm { s } _ { 2 } = 0.0782 \mathrm { fl } \mathrm { oz } \right)$

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Test statistic:
Upper critical F value...

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Question 93

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Determine whether the samples are independent or dependent. -The effectiveness of a headache medicine is tested by measuring the intensity of a headache in patients before and after drug treatment. The data consist of before and after intensities for each patient.

Determine whether the samples are independent or dependent. -A researcher was interested in comparing the salaries of female and male employees at a particular company. Independent simple random samples of 8 female employees and 15 male employees yielded the following weekly salaries (in dollars). $\begin{array} { | r | r r | } \hline \text { Female } & \text { Male } & \\\hline 495 & 722 & 518 \\760 & 562 & 904 \\556 & 880 & 1150 \\904 & 520 & 805 \\520 & 500 & 480 \\1005 & 1250 & 970 \\743 & 750 & 605 \\660 & 1640 & \\\hline\end{array}$ Use a 0.05 significance level to test the claim that the mean salary of female employees is less than the mean salary of male employees. Use the traditional method of hypothesis testing. $\text { (Note: } \left. \bar { x } _ { 1 } = \$ 705.375 , \bar { x } _ { 2 } = \$ 817.067 , \mathrm {~s} _ { 1 } = \$ 183.855 , \mathrm {~s} _ { 2 } = \$ 330.146 . \right)$

The two data sets are dependent. Find $\overline { \mathrm { d } }$ to the nearest tenth. - $\begin{array}{l|lllllll}\hline \mathrm{X} & 236 & 190 & 220 & 182 & 253 & 295 & 302 \\\hline \mathrm{Y} & 194 & 153 & 195 & 153 & 235 & 253 & 284\end{array}$

Determine whether the following statement regarding the hypothesis test for two population proportions is true or false: However small the difference between two population proportions, for sufficiently large sample sizes, the null Hypothesis of equal population proportions is likely to be rejected.

Write the word or phrase that best completes each statement or answers the question. Use the traditional method to test the given hypothesis. Assume that the samples are independent and that they have been randomly selected - $\mathrm { x } _ { 1 } = 15 , \mathrm { n } _ { 1 } = 50 \text { and } \mathrm { x } _ { 2 } = 23 , \mathrm { n } _ { 2 }$ = 60; Construct a 90% confidence interval for the difference between population proportions p1 - p2.

Assume that the following confidence interval for the difference in the mean time (in minutes) for male students to complete a statistics test (sample 1) and the mean time for female students to complete a statistics test (sample2) was constructed using independent simple random samples. $- 0.2 \text { minutes } < \mu _ { 1 } - \mu _ { 2 } < 2.7 \text { minutes }$ What does The confidence interval suggest about the difference in length between male and female test completion times?

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. -When 25 randomly selected customers enter any one of several waiting lines, their waiting times have a standard deviation of 5.35 minutes. When 16 randomly selected customers enter a single main waiting line, their waiting times have a standard deviation of 2.2 minutes. Use a 0.05 significance level to test the claim that there is more variation in the waiting times when several lines are used.

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. -A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their competition scores were recorded before and after the training. The results are shown below. $\begin{array} { c | c c c c c c c } \text { Subject } & \mathrm { A } & \mathrm { B } & \mathrm { C } & \mathrm { D } & \mathrm { E } & \mathrm { F } & \mathrm { G } \\\hline \text { Before } & 9.5 & 9.4 & 9.6 & 9.5 & 9.5 & 9.6 & 9.7 \\\hline \text { After } & 9.6 & 9.6 & 9.6 & 9.4 & 9.6 & 9.9 & 9.5\end{array}$ Using a 0.01 level of significance, test the claim that the training technique is effective in raising the gymnasts' scores.

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. -Use the summary statistics below to test the claim that the samples come from populations with different variances. Use a significance level of 0.05. $\begin{array} { l l l } \frac { \text { Sample A } } { \mathrm { n } = 28 } & & \frac { \text { Sample B } } { \mathrm { n } = 41 } \\\overline { \mathrm { x } } 1 \mathrm { 1 } = 19.2 & & \overline { \mathrm { x } } _ { 2 } = 23.7 \\\mathrm {~s} = 4.78 & & \mathrm {~s} = 5.93\end{array}$

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Independent samples from two different populations yield the following data. $\bar { x } _ { 1 } = 958 , \bar { x } _ { 2 } = 157 , s _ { 1 } = 77 , s _ { 2 } = 88$ . The sample size is 478 for both samples. Find the $85 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ .

Determine whether the samples are dependent or independent. The effectiveness of a new headache medicine is tested by measuring the amount of time before the headache is cured for patients who use the medicine and Another group of patients who use a placebo drug.

State what the given confidence interval suggests about the two population means. -A researcher wishes to determine whether the blood pressure of vegetarians is, on average, lower than the blood pressure of nonvegetarians. Independent simple random samples of 85 vegetarians and 75 nonvegetarians yielded the following sample statistics for systolic blood pressure: $\begin{array} { | l l | } \hline \text { Vegetarians } & \text { Nonvegetarians } \\ \overline { \mathrm { n } _ { 1 } } = 85 & \overline { \mathrm { n } _ { 2 } } = 75 \\\overline { \mathrm { x } _ { 1 } } = 124.1 \mathrm { mmHg } & \overline { \mathrm { x } _ { 2 } } = 138.7 \mathrm { mmHg } \\\mathrm { s } _ { 1 } = 38.7 \mathrm { mmHg } & \mathrm { s } _ { 2 } = 39.2 \mathrm { mmHg } \\\hline\end{array}$ Use a significance level of 0.01 to test the claim that the mean systolic blood pressure for vegetarians is lower than the mean systolic blood pressure for nonvegetarians. Use the P-value method of hypothesis testing.

State what the given confidence interval suggests about the two population means. -A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not Exercise regularly and 12 people who do exercise regularly. The resting pulse rate (in beats per minute) of each Person was recorded. The summary statistics are as follows. $\begin{array} { | r | r | } \hline \text { Do Not Exercise } & \text { Do Exercise } \\\hline \bar { x } _ { 1 } = 72.5 \text { beats } / \mathrm { min } & \bar { x } _ { 2 } = 68.5 \text { beats } / \mathrm { min } \\\mathrm { s } _ { 1 } = 10.4 \text { beats } / \mathrm { min } & \mathrm { s } _ { 2 } = 8.9 \text { beats } / \mathrm { min } \\\mathrm { n } _ { 1 } = 16 & \mathrm { n } _ { 2 } = 12 \\\hline\end{array}$ Construct a 90% confidence interval for the difference between the mean pulse rate of people who do not Exercise regularly and the mean pulse rate of people who exercise regularly.

Determine the decision criterion for rejecting the null hypothesis in the given hypothesis test; i.e., describe the values of the test statistic that would result in rejection of the null hypothesis. -Suppose you wish to test the claim that $\mu _ { \mathrm { d } }$ , the mean value of the differences d for a population of paired data, is different from 0 . Given a sample of $n = 23$ and a significance level of $\alpha = 0.05$ , what criterion would be used for rejecting the null hypothesis?

State what the given confidence interval suggests about the two population means. -A researcher was interested in comparing the response times of two different cab companies. Companies A and B were each called at 50 randomly selected times. The calls to company A were made independently of the calls to company B. The response times were recorded and the summary statistics were as follows: $\begin{array} { l c c } & \text { Company A } & \text { Company B } \\\hline \text { Mean response time } & 7.6 \text { mins } & 6.9 \text { mins } \\\text { Standard deviation } & 1.4 \text { mins } & 1.7 \text { mins }\end{array}$ Use a 0.02 significance level to test the claim that the mean response time for company A differs from the mean response time for company B. Use the P-value method of hypothesis testing.

Assume that you plan to use a significance level of $\alpha = 0.05$ test the claim that p $\mathrm { p } _ { 1 } = \mathrm { p } _ { 2 }$ . Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test. -In a vote on the Clean Water bill, 46% of the 205 Democrats voted for the bill while 48% of the 230 Republicans voted for it.

Solve the problem. -The sample size needed to estimate the difference between two population proportions to within a margin of error $E$ with a confidence level of 1 - $\alpha$ can be found as follows: in the expression $\mathrm { E } = \mathrm { z } _ { \alpha / 2 } \sqrt { \frac { \mathrm { p } 1 \mathrm { q } _ { 1 } } { \mathrm { n } _ { 1 } } + \frac { \mathrm { p } \mathrm { p } _ { 2 } } { \mathrm { n } _ { 2 } } }$ replace $\mathrm { n } _ { 1 }$ and $\mathrm { n } _ { 2 }$ by $\mathrm { n }$ (assuming both samples have the same size) and replace each of $\mathrm { p } _ { 1 } , \mathrm { q } _ { 1 } , \mathrm { p } _ { 2 }$ , and $\mathrm { q } _ { 2 }$ by $0.5$ (because their values are not known) . Then solve for $\mathrm { n }$ . Use this approach to find the size of each sample if you want to estimate the difference between the proportions of men and women who plan to vote in the next presidential election. Assume that you want 99% confidence That your error is no more than 0.02.

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. -A researcher wishes to determine whether listening to music affects students' performance on memory test. He randomly selects 50 students and has each student perform a memory test once while listening to music and once without listening to music. He obtains the mean and standard deviation of the 50 "with music" scores and obtains the mean and standard deviation of the 50 "without music scores". He then performs a hypothesis test for two means assuming large and independent samples. Is this approach appropriate? If not, how would you proceed.