91Ó°ÊÓ

(a)

Given claim: Differs

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 an equality.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{98.1 - 129.2 - 0}}{{\sqrt {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} }} \approx - 2.836\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {14.{2^2}/11} \right)}^2}}}{{11 - 1}} + \frac{{{{\left( {39.{1^2}/15} \right)}^2}}}{{15 - 1}}}} \approx 18\)

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 d f=18 :

\(0.01 = 2 \times 0.005 < P < 2 \times 0.01 = 0.02\)

We then note that the given t-value is not correct

(b)

Given claim: 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 an equality.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = - 25\\{H_a}:{\mu _1} - {\mu _2} < - 25\end{array}\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{98.1 - 129.2 - ( - 25)}}{{\sqrt {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} }} \approx - 0.556\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{14.{2^2}}}{{11}} + \frac{{39.{1^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {14.{2^2}/11} \right)}^2}}}{{11 - 1}} + \frac{{{{\left( {39.{1^2}/15} \right)}^2}}}{{15 - 1}}}} \approx 18\)

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 d f=18 :

\(P > 2 \times 0.10 = 0.20\)

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

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

There is not sufficient evidence to support the claim that the true average strength for males exceeds that for females by more than 25 N.