Question

Heights Use a 0.01 significance level with the sample data from Exercise 3 to test the claim that women have heights with a mean that is less than the mean height of men.

Short answer

There is enough evidence to conclude that the mean height of women is less than the mean height of men.

Step-by-step solution

Given information

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

Step 2:Hypotheses

It is claimed that the mean height of women is less than the mean height of men.

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

Null Hypothesis: The mean height of women is equal to the mean height of men.

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

Alternative Hypothesis: The mean height of women is less than the mean height of men.

\({H_1}:{\mu _1} < {\mu _2}\)

The test is left-tailed.

Important values

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:

\(\begin{aligned} {{\bar x}_1} &= \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_{1i}}} }}{{{n_1}}}\\ &= \frac{{160.3 + 167.7 + ...... + 171.1}}{{10}}\\ &= 162.35\end{aligned}\)

The sample mean height of men is computed below:

\(\begin{aligned} {{\bar x}_2} &= \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_{2i}}} }}{{{n_2}}}\\ &= \frac{{190.3 + 169.8 + ...... + 181.3}}{{10}}\\ &= 178.77\end{aligned}\)

The sample variance of heights of women is computed below:

\(\begin{aligned} s_1^2 &= \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ &= \frac{{{{\left( {160.3 - 162.35} \right)}^2} + {{\left( {167.7 - 162.35} \right)}^2} + ....... + {{\left( {166.9 - 162.35} \right)}^2}}}{{10 - 1}}\\ &= 140.35\end{aligned}\)

The sample variance of heights of men is computed below:

\(\begin{aligned} s_2^2 &= \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ &= \frac{{{{\left( {190.3 - 178.77} \right)}^2} + {{\left( {169.8 - 178.77} \right)}^2} + ....... + {{\left( {181.3 - 178.77} \right)}^2}}}{{10 - 1}}\\ &= 28.11\end{aligned}\)

The following values are obtained from the previous exercise:

\(\begin{array}{l}{{\bar x}_1} = 162.35\\{{\bar x}_2} = 178.77\\{n_1} = 10\\{n_2} = 10\\s_1^2 = 140.35\\s_2^2 = 28.11\end{array}\)

Calculate test statistic, critical value and p-value

The test statistic value is computed as follows:

\(\begin{aligned} t &= \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\;\;\;{\rm{where}}\;{\mu _1} - {\mu _2}\;is\;0\\ &= \frac{{\left( {162.35 - 178.77} \right) - 0}}{{\sqrt {\frac{{140.35}}{{10}} + \frac{{28.11}}{{10}}} }}\\ &= - 4.001\end{aligned}\)

The confidence level is equal to 95%.Thus, the corresponding level of significance is equal to 0.05.

The degrees of freedom are computed below:

\(\begin{aligned} df &= {\rm{smaller}}\;of\;\left( {{n_1} - 1} \right)\;{\rm{and}}\;\left( {{n_2} - 1} \right)\\ &= {\rm{smaller}}\;of\;\left( {10 - 1} \right)\;{\rm{and}}\;\left( {10 - 1} \right)\\ &= 9\end{aligned}\)

Referring to t-table, the critical value of t for\(\alpha = 0.01\)and 9 degrees of freedom for a left-tailed test is equal to –2.8214.

Referring to t-table, the corresponding p-value is equal to 0.0016 which is obtained by using the test statistic (–4.001).

Since the test statistic (–4.001) is less than the critical value (–2.8214) and the p-value is less than 0.01, the null hypothesis is rejected.