A hospital wishes to justify the benefits of nutrition progr

A hospital wishes to justify the benefits of nutrition programs for pregnant women using birth weight data from newborns. The hospital hopes to show that the mean birth weight for newborns from mothers who complete the program is higher than the birth weight for newborns from mothers who do not complete the program. A group of 16 pregnant women were randomly divided into two groups; the first group received the nutrition program and the second group did not receive the program. The resulting weights (in grams) of the newborn babies from each group are shown below. Assume normality.
Group 1,Group 2
2358,2548
2382,2574
2683,2439
2489,2499
2640,2579
2572,2361
2713,2567
2676,2601
2666,2344

b)This question is asking you to do a test on the equality of variance. What is the P-value for the test where the null is that the variability in birth weight is the same regardless of whether the mother completes the program and the alternative is two sided? Give your answer to four decimal places.

c) For the primary test of interest specified in Part a, use the test that assumes equal variances since the P-value in Part b is not significant. What is the test statistic? Give your answer to four decimal places.

d) What is the P-value associated with the test statistic? Give your answer to four decimal places.

e) What is the appropriate conclusion for the hospital using a 0.05 level of significance?
Fail to reject the claim that the mean birth weight with the program is equal to the mean birth weight without the program because the P-value is less than 0.05.

Reject the claim that the mean birth weight with the program is higher than the mean birth weight without the program because the P-value is less than 0.05.  

Conclude that the mean birth weight with the program is higher than the mean birth weight without the program because the P-value is less than 0.05.

Fail to reject the claim that the mean birth weight with the program is equal to the mean birth weight without the program because the P-value is greater than 0.05.

Solution