Variation of Heights Use the sample data given in Exercise 3 “Heights” and test the claim that women and men have heights with the same variation. Use a 0.05 significance level.

Two samples of heights of women and men (in cm) are tabulated.

It is claimed that women and men have heights with the same variation.

Corresponding to the given claim, the following hypotheses are set up:

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

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

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

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

The test is two-tailed.

If the test statistic does not lie between the two critical values, the null hypothesis is rejected.

The sample size for the heights of women is equal to\({n_1} = 10\).

The sample size for the heights of men is equal to\({n_2} = 10\).

The sample mean height of women is computed below:

\(\begin{aligned} {{\bar x}_1} &= \frac{{160.3 + 167.7 + ...... + 171.1}}{{10}}\\ &= 162.35\end{aligned}\)

The sample mean height of men is computed below:

\(\begin{aligned} {{\bar x}_2} &= \frac{{190.3 + 169.8 + ...... + 181.3}}{{10}}\\ &= 178.77\end{aligned}\)

The sample variance of heights of women is computed below:

\(\begin{aligned} s_1^2 &= \frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}\\ &= \frac{{{{\left( {160.3 - 162.35} \right)}^2} + {{\left( {167.7 - 162.35} \right)}^2} + ....... + {{\left( {166.9 - 162.35} \right)}^2}}}{{10 - 1}}\\ &= 140.35\end{aligned}\)

The sample variance of heights of men is computed below:

\(\begin{aligned} s_2^2 &= \frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}\\ &= \frac{{{{\left( {190.3 - 178.77} \right)}^2} + {{\left( {169.8 - 178.77} \right)}^2} + ....... + {{\left( {181.3 - 178.77} \right)}^2}}}{{10 - 1}}\\ &= 28.11\end{aligned}\)

The value of the test statistic is computed as shown:

\(\begin{aligned} F &= \frac{{s_1^2}}{{s_2^2}}\\ &= \frac{{140.35}}{{28.11}}\\ &= 4.99\end{aligned}\)

The degrees of freedom are computed below:

\(\begin{aligned} df &= \left( {{n_1} - 1,{n_2} - 1} \right)\\ &= \left( {10 - 1,10 - 1} \right)\\ &= \left( {9,9} \right)\end{aligned}\)

The critical values of F at\(\alpha = 0.05\)and degrees of freedom equal to (9,9) (for numerator and denominator)are 0.2484 and 4.026.

The corresponding p-value is equal to 0.0252.

Since the absolute value of the test statistic (4.99) does not lie between the critical values and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that women and men do not have the same variation in their heights. In other words, there is enough evidence to warrant rejection of the claim that women and men have heights with the same variation.