91影视

Given:

\(\begin{array}{l}{n_1} = 10\\{n_2} = 5\\\alpha = 0.10\end{array}\)

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

\(\begin{array}{l}{{\bar x}_1} = \frac{{29 + 34 + 33 + \ldots + 34 + 32 + 27}}{{10}} \approx 30.7\\{{\bar x}_2} = \frac{{18 + 15 + 23 + 13 + 12}}{5} \approx 16.2\end{array}\)

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:

\(\begin{array}{l}{s_1} = \sqrt {\frac{{{{(29 - 30.7)}^2} + \ldots . + {{(27 - 30.7)}^2}}}{{10 - 1}}} \approx 2.7508\\{s_2} = \sqrt {\frac{{{{(18 - 16.2)}^2} + \ldots . + {{(12 - 16.2)}^2}}}{{5 - 1}}} \approx 4.4385\end{array}\)

Given claim: more than \(10\)

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.

\(\begin{array}{l}{H_0}:{\mu _1} - {\mu _2} = 10\\{H_a}:{\mu _1} - {\mu _2} > 10\end{array}\)

Determine the test statistic:

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{30.7 - 16.2}}{{\sqrt {\frac{{{{2.7508}^2}}}{{10}} + \frac{{{{4.4385}^2}}}{5}} }} \approx 6.691\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{2.7508}^2}}}{{10}} + \frac{{{{4.4385}^2}}}{5}} \right)}^2}}}{{\frac{{{{\left( {{{2.7508}^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {{{4.4385}^2}/5} \right)}^2}}}{{5 - 1}}}} \approx 5\)

The \({\rm{P}}\)-value is the probability of obtaining the value of the test statistic, or a value more extreme. The \({\rm{P}}\)-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 \(df = 5\) :

\(0.0005 < P < 0.001\)

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

\(P < 0.10 \Rightarrow {\rm{ Reject }}{H_0}\)

There is sufficient evidence to support the claim that the true average maximum lean angle for older females is more than degrees smaller than it is for younger females.