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(a)

Given:

\(\begin{array}{l}\bar x = 325.73\\\\s = 34.97\\\\ n = 6\end{array}\)

I will estimate using a 95% confidence interval. Other confidence levels can be obtained similarly.

\(c = 95\% = 0.95\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 6 - 1 = 5\)and in the column with\(\alpha = 1 - c/2 = 0.025\)in the table of the Student's T distribution:

\({t_\alpha } = 2.571\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 2.571 \times \frac{{34.97}}{{\sqrt 6 }} \approx 36.7047\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}\bar x - E = 325.73 - 36.7047 = 289.0253\\\bar x + E = 325.73 + 36.7047 = 362.4347\end{array}\)

We are 95 % confident that the true average breaking force in a dry medium at \(3{7^\circ }\) is between 289.0253 and 362.4347

(b)

Given:

\(\begin{array}{l}{{\bar x}_1} = 325.73\\{s_1} = 34.97\\ {{\bar x}_2} = 306.09\\{s_2} = 41.97\\{n_1} = {n_2} = 6\end{array}\)

I will estimate using a 95% confidence interval. Other confidence levels can be obtained similarly.

\(c = 95\% = 0.95\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{34.9{7^2}}}{6} + \frac{{41.9{7^2}}}{6}} \right)}^2}}}{{\frac{{{{\left( {34.9{7^2}/6} \right)}^2}}}{{6 - 1}} + \frac{{{{\left( {41.9{7^2}/6} \right)}^2}}}{{6 - 1}}}} \approx 9 \)

Determine the t-value by looking in the row starting with degrees of freedom $d f=9$ and in the column with\(\alpha = 1 - c/2 = 0.025\)in the Student's t distribution table in the appendix:

\({t_{\alpha /2}} = 2.262\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.262 \times \sqrt {\frac{{34.9{7^2}}}{6} + \frac{{41.9{7^2}}}{6}} \approx 50.4480\)

The endpoints of the confidence interval for\({\mu _1} - {\mu _2}\)are:

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (325.73 - 306.09) - 50.4480 = 19.64 - 50.4480 = - 30.8080 \\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (325.73 - 306.09) + 50.4480 = 19.64 + 50.4480 = 70.0880\end{array}\)

(c)

Given:

\(\begin{array}{l}{{\bar x}_1} = 325.73\\{s_1} = 34.97\\{{\bar x}_2} = 170.60\\{s_2} = 39.08\\{n_1} = {n_2} = 6\end{array}\)

Let us assume:\(\alpha = 0.01\)

Given claim: Exceeds by more than 100

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = 100\\{H_a}:{\mu _1} - {\mu _2} > 100\end{array}\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{34.9{7^2}}}{6} + \frac{{39.0{8^2}}}{6}} \right)}^2}}}{{\frac{{{{\left( {34.9{7^2}/6} \right)}^2}}}{{6 - 1}} + \frac{{{{\left( {39.0{8^2}/6} \right)}^2}}}{{6 - 1}}}} \approx 9\)

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{325.73 - 170.60 - 100}}{{\sqrt {\frac{{34.9{7^2}}}{6} + \frac{{39.0{8^2}}}{6}} }} \approx 2.575\)

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=9 :

\(0.01 < P < 0.025\)

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

\(P < 0.05 \Rightarrow Reject {H_0}\)

There is sufficient evidence to support the claim that the true average force in a dry medium at the higher temperature exceeds that at the lower temperature by more than 100 N.