Question

A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.01 significance level for both parts.			Men	Women
		muμ	μ1	μ2
		n	11	59
		x overbarx	97.59 °F	97.36 °F
		s	0.96 °F	0.72 °F

a. Test the claim that men have a higher mean body temperature than women.

What are the null and alternative​ hypotheses?

A. H0​: μ1 ≠ μ2

H1​: μ1 < μ2

B. H0​: μ1 = μ2

H1​: μ1 ≠ μ2

C. H0​: μ1 = μ2

H1​: μ1 > μ2

D. H0​: μ1g ≥ μ2

H1​: μ1 < μ2

The test​ statistic, t, is ??? ​(Round to two decimal places as​ needed.)

The​ P-value is ??? ​(Round to three decimal places as​ needed.)

State the conclusion for the test.

A. Fail to rejectFail to reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women.

B. Fail to rejectFail to reject the null hypothesis. There is notis not sufficient evidence to support the claim that men have a higher mean body temperature than women.

C. Reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women.

D. Reject the null hypothesis. There is notis not sufficient evidence to support the claim that men have a higher mean body temperature than women.

b. Construct a confidence interval suitable for testing the claim that men have a higher mean body temperature than women.

???<mμ1−μ2<???

​(Round to three decimal places as​ needed.)

Does the confidence interval support the conclusion found with the hypothesis​ test?

▼??(Yes, No,) because the confidence interval contains ▼(zero., only negative values.,only positive values.)

2.

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.01significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm	147	136	141	133	132
Left arm	172	168	191	155	151

In this​ example, μd is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis​ test?

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a

0.010.01

significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm	147147	136136	141141	133133	132132
Left arm	172172	168168	191191	155155	151151

In this​ example,

mu Subscript dμd

is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis​ test?

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a

0.010.01

significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm	147147	136136	141141	133133	132132
Left arm	172172	168168	191191	155155	151151

In this​ example,

mu Subscript dμd

is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis​ test?

A. H0​: μd ≠ 0

H1​: μd > 0

B. H0​: μd = 0

H1​: μd < 0

C.H0​:μd = 0

H1​: dμ ≠ 0

D. H0​:μd ≠ 0

H1​:μd = 0

Identify the test statistic.

t = ??? ​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-value =??? ​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

Since the​ P-value is ▼(less, greater) than the significance​ level, ▼(fail to reject, reject) the null hypothesis. There ▼(is not, is) sufficient evidence to support the claim of a difference in measurements between the two arms.

Answer 1

1 HYpothesis Testing

The test hypothesis is

This is a one-sided test because the alternative hypothesis is formulated to detect the difference from the hypothesized mean on the upper side
Now, the value of test static can be found out by following formula:

.

Degrees of freedom on the t-test statistic are n1 + n2 - 2 = 11 + 59 - 2 = 68
This implies that

OPTION B
Fail to reject the null hypothesis. There is notis not sufficient evidence to support the claim that men have a higher mean body temperature than women.

CONFIDENCE INTERVAL

Confidence interval(in %) = 99

Given, POpulation variances are not equal
First we calculate degree of freedom

Required confidence interval = (0.23 - 0.763, 0.23 + 0.763)
Required confidence interval = (-0.533, 0.993)

YES
the confidence interval support the conclusion found with the hypothesis​ test because the confidence interval contains ZERO