For the problem show the calculator command you are using. Round Pvalue to three decimal places. Indicate the specific values that you are typing in the calculator foe each problem. Interpretations should include the terms "sufficient evidence" or "insufficient evidence" in the context of the problem.

Breast feeding sometimes results is a temporary loss of bone mass, as calcium is depleted in the mother's body to provide for milk production. Data on total body bone mineral content (in grams) for a sample of mothers both during breast feeding (B) and in the post weaning (P) are given below. Do the data suggest that actual mean bone mineral content during post weaning is greater that during breast feeding by more than 25 grams? Use a significance level of 0.05. (Hint: Us post minus during for the set up. Then instead of a difference of 0 for the hypotheses you are looking at a difference of 25).

Mother 1 2 3 4 5 6 7 8 9 10

P 2126. 2885. 2895. 1942 1750. 2184 2164 2626. 2006. 2627.

B 1928. 2549. 2825 1924 1628 2175. 2144. 2621. 1843. 2541

there is no enough evidence to prove that actual mean bone mineral content during post weaning is greater that during breast feeding by more than 25 grams

Let group 1 be data set for during breastfeeding and group 2 be data set for after breastfeeding

Let H0 be the null hypothesis and H1 be the alternative hypothesis.

H0 : u1-u2 = 25

H1 : u1 < u2

The sample mean and variance of the data set was gotten via Excel.

Kindly find attachment below for more details.

x1 = 2320.5, n1 = 10, s1² = 164953.39

x2 = 2217.8, n2 = 10, s2² = 156814.84

Test statistics is a t test given below as

t = (x1 - x2)- (u1-u2)/√s1²/n1 + s2²/n2

t = (2320.5 - 2217.8) - 25/√164953.39/10 + 156814.84/10

t = 77.5/179.37899

t = 0.4321

Test statistics is 0.4321

To make a conclusion, we get the p value of the corresponding test statistics and compare with the level of significance.

If p > 0.05, we accept the null hypothesis.

If p < 0.05, we reject the null hypothesis.

The p value of this test statistics can be gotten either from a t distribution table or any online p value calculator.

Link to calculator below

https://www.socscistatistics.com/pvalues/tdistribution.aspx

From my calculator, the p value is 0.338

Since 0.338 is greater than 0.05, we accept the null hypothesis

In conclusion, there is no enough evidence to prove that actual mean bone mineral content during post weaning is greater that during breast feeding by more than 25 grams