Question

u 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 0.05 significance level for both parts Men 11 11 97.76°F 0.81°F Women 2 59 97.45°F 0.71°F S a. Test the claim that men have a higher mean body temperature than women What are the null and alternative hypotheses? OB. Ho H1 = H2 H:H *12 O A. Ho 1122 на: 1 <p2 0C. Ho H = 12 H:14 > H2 OD. Horu * 42 H: HI2 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. O A. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women. O B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women O C. Reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women. OD. Reject the null hypothesis. There is 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. ]<H-H2O (Round to three decimal places as needed.) Does the confidence interval support the conclusion found with the hypothesis test? because the confidence interval contains

Answer 1

• Here, we are to test if the mean of men is greater than the mean of female. That is, we are to test:
  $H_0:\mu_1=\mu_2 \text{ v/s }H_1:\mu_1>\mu_2$
  That is, option C) is correct.
  
• The test statistic is given by:
  $T=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$
  Given Information:
  $n_1=11$
  $n_2=59$
  $\overline{x _1}=97.76$
  $\overline{x _2}=97.45$
  $s_1=0.81$
  $s_2=0.71$
  
  Hence,
  $T=\frac{97.76-97.45}{\sqrt{\frac{0.81^2}{11}+\frac{0.71^2}{59}}}=1.187142545$
  
  
• P-value
  The p-value is given by:
  $P(t_{n_1+n_2-2}>T)$
  
  
  
  $=1-0.8803$
  $=0.1197$
  
• Conclusion
  Since the p-value is not less than 0.05, we fail to reject the Null hypothesis. Hence, the correct option is B). That is,
  Fail to reject the Null Hypothesis. There is not sufficient evidence to support the claim that men has a higher mean body temperature than women .
  
• Confidence Interval
  The 95% Confidence Interval is given by:
  $(\overline{x_1}-\overline{x_2}-\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}*t_{n_1+n_2-2;0.025},\overline{x_1}-\overline{x_2}+\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}*t_{n_1+n_2-2;0.025})$
  Putting in the given values, we get,
  $\tiny (97.76-97.45-\sqrt{\frac{0.45^2}{11}+\frac{0.71^2}{59}}*1.995,97.76-97.45+\sqrt{\frac{0.45^2}{11}+\frac{0.71^2}{59}}*1.995)$
  
  
  
• Yes, the Confidence Interval suppurts our claim. This is because, the 95% Confidence Interval contains the value under the Null hypothesis, 0. Hence, the correct option is:
  Yes, because the confidence interval contains the claim under Null,0 .
  

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