Question

40. Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk produc tion. The Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes" (Amer. J. of Clinical Nutr. 2004: 1322-1326) gave the following data on total body bone mineral content (TBBMC) (g) for a sample both during lactation (L) and in the postweaning period (P) paper "Bone Mass Is Recovered from Subject 1 2 34 56 78 910 L 1928 2549 2825 1924 1628 2175 2114 2621 1843 2541 P 2126 2885 2895 1942 1750 2184 2164 2626 2006 2627 a. Does the data suggest that true average total body bone mineral content during postweaning exceeds that during lactation by more than 25 g? State and test the appropriate hypotheses using a significance level of .05. [Note: The appropriate normal proba- bility plot shows some curvature but not enough to cast substantial doubt on a normality assumption.] b. Calculate an upper confidence bound using a 95% average difference confidence level for the true between TBBMC during postweaning and during lactation. c. Does the (incorrect) use of the two-sample t test to test the hypotheses suggested in (a) lead to the same conclusion that you obtained there? Explain.

MAKE SURE YOU FULLY EXPLAIN PART (c). The mentioning of "incorrect" use of two sample t test in (c) means (a) would be some other t test.

Answer 1

a)

L	P	diff
1928	2126	-198
2549	2885	-336
2825	2895	-70
1924	1942	-18
1628	1750	-122
2175	2184	-9
2114	2164	-50
2621	2626	-5
1843	2006	-163
2541	2627	-86
	mean	-105.7
	sd	103.845

The differences for the sample are:

-198 -336 -70 -18 -122 -9 -50 -5 -163 -86

with mean d¯ = −105.7 and standard deviation sD = 103.8

TS = (-105.7 - 25)/(103.8/sqrt(10))

= -3.98178

p-value = P(t < -3.98178)

= 0.0016

p-value < alpha

hence we reject the null hypothesis

b)

t-critical = 1.8331

hence

upper confidence bound = (-105.7 + 1.833 * 103.845/sqrt(10))

= -45.506

c)

Using two-sample t-test

Using Excel

t-Test: Two-Sample Assuming Equal Variances
	L	P
Mean	2214.8	2320.5
Variance	157396.8444	164953.3889
Observations	10	10
Pooled Variance	161175.1167
Hypothesized Mean Difference	25
df	18
t Stat	-0.727966834
P(T<=t) one-tail	0.237998918
t Critical one-tail	1.734063607
P(T<=t) two-tail	0.475997836
t Critical two-tail	2.10092204

p-value for 1-tail = 0.2380

p-value > 0.05

hence we fail to reject the null hypothesis

observe that in a) part we rejected the null hypothesis ,here we are not

a) is actually paired t-test here we take the difference between bone mineral content during Postwearing and during Lactation , whereas c) is 2-sample independent test

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