Question

Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk production. The paper "Bone Mass Is Recovered from 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).

\(\begin{array}{*{20}{c}}{Subject}&{}&{}&{}&{}&{}&{}&{}&{}&{}\\1&2&3&4&5&6&7&8&9&{10}\\{1928}&{2549}&{2825}&{1924}&{1628}&{2175}&{2114}&{2621}&{1843}&{2541}\\{2126}&{2885}&{2895}&{1942}&{1750}&{2184}&{2164}&{2626}&{2006}&{2627}\end{array}\)

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 probability plot shows some curvature but not enough to cast substantial doubt on a normality assumption.)

b. Calculate an upper confidence bound using a\(95\% \)confidence level for the true average difference 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

Short answer

(a) There is sufficient evidence to support the claim that the true average total body bone mineral content during postweaning exceeds that during lactation by more than\(25\;{\rm{g}}\).

(b) \( - 45.5067\)

(c) Not the same conclusion

Step-by-step solution

A)Step 1: Find the difference

Determine the difference in value of each pair.

Sample 1	Sample 2	Difference D
1928	2126	-198
2549	2885	-335
2825	2895	-70
1924	1942	-18
1628	1750	-122
2175	2184	-9
2114	2164	-50
2621	2626	-5
1843	2006	-163
2541	26277	-86

Find the value of test statistics

Given claim: difference between \({\rm{P}}\) and \({\rm{L}}\) is more than \(25.\)

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _d} = - 25\\{H_a}:{\mu _d} < - 25\end{array}\)

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 198 - 336 - 70 + \ldots - 5 - 163 - 86}}{{10}} \approx - 105.7\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 198 - ( - 105.7))}^2} + \ldots + {{(86 - ( - 105.7))}^2}}}{{10 - 1}}} \approx 103.8450\)

Determine the value of the test statistic:

\(t = \frac{{\bar d - d}}{{{s_d}/\sqrt n }} = \frac{{ - 105.7 - ( - 25)}}{{103.8450/\sqrt {10} }} \approx - 2.457\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The Pvalue is the number (or interval) in the column title of the Student's T distribution in the appendix containing the t-value in the row\(df = n - 1 = \$ \$ 10 - 1 = 9\) :

\(0.01 < P < 0.025\)

If the P-value is less than the significance level, reject the null hypothesis.

\(P < 0.05 \Rightarrow {\mathop{\rm Reject}\nolimits} {H_0}\)

There is sufficient evidence to support the claim that the true average total body bone mineral content during postweaning exceeds that during lactation by more than\(25\;{\rm{g}}\)

b)Step 3: Find the endpoint of confidence interval

\(c = 95\% = 0.95\)

Determine the\({t_{\alpha /2}}\)using the Student's T distribution table in the appendix with\(df = n - 1 = 10 - 1 = 9\):

\({t_{0.05}} = 1.833\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 1.833 \cdot \frac{{103.8450}}{{\sqrt {10} }} \approx 60.1933\)

The endpoints of the confidence interval for\({\mu _d}\)are:

\(\bar d + E = - 105.7 + 60.1933 = - 45.5067\)

Thus the upper confidence bound is\( - 45.5067\)

c)Step 4:

Find the standard deviaton

The mean is the sum of all values divided by the number of values:

\(\begin{array}{l}{{\bar x}_1} = \frac{{1928 + 2549 + 2825 + \ldots + 2621 + 1843 + 2541}}{{10}} = 2214.8\\{{\bar x}_2} = \frac{{2126 + 2885 + 2895 + \ldots + 2626 + 2006 + 2627}}{{10}} = 2320.5\end{array}\)

The variance is the sum of squared deviations from the mean divided by\(n - 1\). The standard deviation is the square root of the variance:

\(\begin{array}{l}{s_1} = \sqrt {\frac{{{{(1928 - 2214.8)}^2} + \ldots . + {{(2541 - 2214.8)}^2}}}{{10 - 1}}} \approx 396.7327\\{s_2} = \sqrt {\frac{{{{(2126 - 2320.5)}^2} + \ldots .. + {{(2627 - 2320.5)}^2}}}{{10 - 1}}} \approx 406.1445\end{array}\)

Find the test statisitc

Determine the test statistic:

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2} - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{2214.8 - 2320.5 - ( - 25)}}{{\sqrt {\frac{{{{396.7327}^2}}}{{10}} + \frac{{{{406.1445}^2}}}{{10}}} }} \approx - 0.449\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{396.7327}^2}}}{{10}} + \frac{{{{406.1445}^2}}}{{10}}} \right)}^2}}}{{\frac{{{{\left( {{{396.7327}^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {{{406.1445}^2}/10} \right)}^2}}}{{10 - 1}}}} \approx 17\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's\(T\)distribution in the appendix containing the t-value in the row\(df = 17\):

\(P > 0.10\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P > 0.05 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

We then note that we made a different conclusion compared to part (a).