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Determine the difference in value of each pair.

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\(25\;{\rm{g}}\)

\(c = 95\% = 0.95\)

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\( - 45.5067\)

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:

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

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).