a. The test statistic, t, is (Round to two decimal places as needed.)b. P-value (Round to three decimal places as needed.)c. Confidence interval (Round to three decimal places as needed.)d. conclusion for hypothesis test MenWomenA study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simplerandom samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a)and (b) below.H2n115997.27°F0.67°F97.59°F0.84°Fa. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women.What are the null and alternative hypotheses?A. Ho: H12 H2B. Ho: H1 =H2H1: H1 H2C. Ho: H1 # H2H1: 41

Answered: H2 A study was done on body…

MATLAB: An Introduction with Applications

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Author:Amos Gilat

Publisher:Amos Gilat

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Question

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

b. P-value (Round to three decimal places as needed.)

c. Confidence interval (Round to three decimal places as needed.)

d. conclusion for hypothesis test

Men
Women
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.
H2
n
11
59
97.27°F
0.67°F
97.59°F
0.84°F
a. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women.
What are the null and alternative hypotheses?
A. Ho: H12 H2
B. Ho: H1 =H2
H1: H1 <H2
H1: H1 > H2
C. Ho: H1 # H2
H1: 41 <H2
Họ: H1 = H2
H1: H1 # H2

Expert Solution

Step 1

(a) Calculation of the test statistic:

Formula:

$\begin{array}{rcl} t & = & \frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{\sqrt{\frac{{s_{1}}^{2}}{n_{1}} + \frac{{s_{2}}^{2}}{n_{2}}}} \\ = & \frac{97.59 - 97.27}{\sqrt{\frac{0 . 84^{2}}{11} + \frac{0 . 67^{2}}{59}}} \\ = & \frac{0.32}{\sqrt{0.064145 + 0.007608}} \\ = & 1.1946 \\ \approx & 1.19 \end{array}$

Thus, the value of the test statistic is 1.19

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