Breast feeding sometimes results in a temporary loss of bone mass as calcium is depleted in the mother's body to provide for milk production. An investigation gave the following data on total body bone mineral content (g) for a sample of mothers both during breast feeding (B) and in the postweaning period (P).

Question

Breast feeding sometimes results in a temporary loss of bone mass as calcium is depleted in the mother's body to provide for milk production. An investigation gave the following data on total body bone mineral content (g) for a sample of mothers both during breast feeding (B) and in the postweaning period (P).

Subject 1 2 3 4 5 6 7 8 9 10
B 2516 2087 2707 2376 2253 2311 2591 2195 2574 2117
P 2560 2164 2736 2465 2357 2505 2617 2218 2773 2195
Do the data suggest that true average total body bone mineral content during postweaning exceeds that during breast feeding by more than 25 g? State and test the appropriate hypotheses using a significance level of .05. (Use a statistical computer package to calculate P-value.)
t = (Round the answer to two decimal places.)
df =
P-value = (Round the answer to three decimal places.)

Answer 1

To determine if the true average total body bone mineral content during postweaning exceeds that during breastfeeding by more than 25 g, we can use a paired t-test.

Hypotheses:
Null Hypothesis (H0): Average total body bone mineral content during postweaning is less than or equal to that during breastfeeding + 25 g (µP ≤ µB + 25)
Alternative Hypothesis (Ha): Average total body bone mineral content during postweaning exceeds that during breastfeeding + 25 g (µP > µB + 25)

We need to calculate the t-value, degrees of freedom (df), and the p-value.

Step 1: Calculate the differences between postweaning (P) and breastfeeding (B) measurements.
Subject 1 2 3 4 5 6 7 8 9 10
B 2516 2087 2707 2376 2253 2311 2591 2195 2574 2117
P 2560 2164 2736 2465 2357 2505 2617 2218 2773 2195
Difference 44 77 29 89 104 194 26 23 199 78

Step 2: Calculate the mean of the differences.
Mean (X̄diff) = (44+77+29+89+104+194+26+23+199+78)/10 = 98.4

Step 3: Calculate the standard deviation of the differences.
Standard deviation (sdiff) = sqrt([(44-98.4)^2 + (77-98.4)^2 + ... + (78-98.4)^2]/(n-1))

Step 4: Calculate the t-value.
t = (X̄diff - 25) / (sdiff / sqrt(n))

We need to use a statistical computer package to calculate the p-value corresponding to the t-value calculated in the previous step.

Once you have the t-value, degrees of freedom (df), and p-value from the statistical computer package, you can compare the p-value to the significance level of 0.05. If the p-value is less than 0.05, we reject the null hypothesis.