Question

A.

9.2.10-T Question Help 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 Inl standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. 5 11 97.52°F 0.79°F 59 97.36"F 0.730F a. Test the claim that men have a higher mean body temperature than women. What are the null and alternative hypotheses? OA. HOE H2H2 Hy: Hy <H2 C. Ho H1 =12 Hp: H1 F12 B. Ho: H1 H2 Hy: HH2 OD. HO: H+H2 Hip H2 The test statistic, t, is (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. A parts 4 remaining Clear All Check Answer

B.

C.

D.

Construct a confidence interval suitable for testing claim that students taking non proctored tests get higher mean score than those taking proctored tests.

___<µ1 - µ2 < ____

Yes/No____ because the confidence interval contains only positive values/only negative values/zero ______.

E.

Construct a confidence interval suitable for testing claim that students taking non proctored tests get higher mean score than those taking proctored tests.

___<µ1 - µ2 < ____

Yes/No____ because the confidence interval contains only positive values/only negative values/zero ______.

PS Can you please help me solve Part A-F. I've been sick all week and I could barely get out of bed. I need your help. I don't know what to do. I can't solve them either so I'm stressing out even more. PLEASE HELP IM BEGGING YOU. MAY GOD BLESS YOUR SOUL. ILL APPRECIATE ALL THE HELP PLEASE.

Answer 1

9.2.10

Claim : Men have higher mean body temperature than women.

The hypothesis are :

H0: $\mu _{1}=\mu _{2}$ v/s H1: $\mu _{1}>\mu _{2}$

Here population standard deviations are not equal. Hence the test statistic is,

$t=\frac{\overline{x}_{1}-\overline{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n1}&plus;\frac{s_{2}^{2}}{n2}}}$

$=\frac{97.52-97.36}{\sqrt{\frac{0.79^{2}}{11}&plus;\frac{0.73^{2}}{59}}}$

= 0.62

Degrees of freedom are given by,

$d.f. = \frac{\left (\frac{s_{1}^{2}}{n1}&plus;\frac{s_{2}^{2}}{n2} \right )^{2}}{\frac{(s_{1}^{2}/n_1)^{2}}{n_{1}-1}&plus;\frac{(s_{2}^{2}/n_2)^{2}}{n_{2}-1}}$

$=\frac{\left ( \frac{0.79^{2}}{11}&plus;\frac{0.73^{2}}{59} \right )^{2}}{\frac{(0.79^{2}/11)^{2}}{11-1}&plus;\frac{(0.73^{2}/59)^{2}}{59-1}}$

= 13.38  

$\approx$ 13

p value is given by,

p value = p ( t > 0.62 )

= 0.273 -------------( using excel formula " =t.dist.rt(0.62, 13)" )

Here p value > $\alpha$ ( 0.05).

Conclusion :

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

b. Here we construct 95% confidence interval.

{ ( $\overline{x}_{1}-\overline{x}_{2}$ ) - E, ( $\overline{x}_{1}-\overline{x}_{2}$ ) - E }

Where,

$E=tc*\sqrt{\frac{s_{1}^{2}}{n1}&plus;\frac{s_{2}^{2}}{n2}}$

c = 0.95, $\alpha$ = 1- c = 1 - 0.95 = 0.05 , df =13

tc = $t_{\alpha /2,df}=t_{0.05/2,13}=2.160$ ------------( using excel formula " =t.inv.2t( 0.05,13 ) " )

Hence the margin of error is given by,

$E=2.160*\sqrt{\frac{0.79^{2}}{11}&plus;\frac{0.73^{2}}{59}}$

= 0.554

Hence the confidence interval is,
{ ( 97.52 -97.36 ) - 0.554 , ( 97.52 - 97.36) + 0.554 }

( 0.16 - 0.554 , 0.16 + 0554 )

( -0.394, 0.714 )

-0.394 < $\mu _{1}-\mu _{2}$ < 0.714

The confidence interval support the conclusion of the test because the confidence interval contains 0.