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The sample standard deviation of the heights of men is equal to 7.10098 cm. The sample standard deviation of the heights of women is 7.48296 cm.

In general, the larger of the two sample variances is denoted by \(s_1^2\) while, the smaller of the two sample variances is denoted by \(s_2^2\).

Here, the values of the sample variances are computed as shown:

\(\begin{array}{c}{s^2}_{women} = {\left( {7.48296} \right)^2}\\ = 55.99469\end{array}\)

\(\begin{array}{c}{s^2}_{men} = {\left( {7.10098} \right)^2}\\ = 50.42392\end{array}\)

It can be observed that the sample variance corresponding to the heights of women is greater than the sample variance corresponding to the heights of men.

Thus, \(s_1^2 = 55.99469\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\) and \(s_2^2 = 50.42392\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\).

b.

To test the significance of the claim that the variances of the heights of women and men are not equal, the hypotheses are formulated as follows:

Null Hypothesis: The variance of the heights of women is equal to the variance of the heights of men.

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternative Hypothesis: The variance of the heights of women is not equal to the variance of the heights of men.

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

c.

The value of the test statistic is computed as follows:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{55.99469}}{{50.42392}}\\ = 1.1105\end{array}\)

Thus, the F-statistic is equal to 1.1105.

d.

The level of significance assumed is equal to 0.05.

The p-value is equal to 0.5225, which is greater than 0.05. So, the null hypothesis is failed to reject.

Therefore, there is not sufficient evidence to support the claim that heights of men and heights of women have different variances