Heights of PresidentsA popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (cm) of presidents along with the heights of their main opponents (from Data Set 15 “Presidents”).

a.Use the sample data with a 0.05 significance level to test the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than 0 cm.

b.Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)?

Height (cm) of President	185	178	175	183	193	173
Height (cm) of Main Opponent	171	180	173	175	188	178

The heights of the presidents and main opponents are recorded. The sample size\(n = 6\). The level of significance \(\alpha = 0.05\). It is claimed that the differences of heights have a mean greater than 0.

The following hypotheses are noted:

Null Hypothesis: The differences in the heights have a mean equal to 0.

Alternative Hypothesis: The differences in the heights have a mean greater than 0.

\[{H_1}:{\mu _d} > 0\]

Since the alternative hypothesis contains greater than symbol, it is a right-tailed test.

The following table shows the differences in heights for each pair:

Themean value of the differences between heights of presidents and the main opponents is:

\(\begin{array}{c}\bar d = \frac{{\sum d }}{n}\\ = \frac{{22}}{6}\\ = 3.67\end{array}\)

Thus, the mean value is 3.67 cm.

Thestandard deviation of the differences between the heights of presidents and the main opponents is:

\[\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {14 - 3.67} \right)}^2} + {{\left( { - 2 - 3.67} \right)}^2} + ...... + {{\left( { - 5 - 3.67} \right)}^2}}}{{6 - 1}}} \\ = 6.89\end{array}\]

Thus, the standard deviation is 6.89 cm.

The test statistic for matched pair is given by,

\[\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{3.67 - 0}}{{\frac{{6.89}}{{\sqrt 6 }}}}\\ = 1.305\end{array}\]

Thus, t=1.305.

The degrees of freedom is equal to

\(\begin{array}{c}df = n - 1\\ = 6 - 1\\ = 5\end{array}\)

Referring to the t-distribution table, use the column for 0.05 (Area in One Tail),and use the row with 5 degrees of freedom to find the critical value.

\(\begin{array}{c}{t_{crit}} = {t_\alpha },df\\ = {t_{0.05}},5\\ = 2.015\end{array}\)

Thus, the critical value is 2.015.

a.

For a right-tailed test, if the calculated test statistic value is greater than the critical value, reject the null hypothesis.

Here, 1.305 < 2.02. Thus, the null hypothesis fails to be rejected.

There is not sufficient evidence to support the claim that the differences between the heights of the presidents and their main opponents have a mean greater than 0.

If the level of significance is 0.05 for one-tailed test, then use 90% confidence level to construct the confidence interval.

The margin of error is computed as follows:

\(\)\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = 2.015 \times \frac{{6.89}}{{\sqrt 6 }}\\ = 5.668\end{array}\)

b.

The formula for the confidence interval is equal to:

\(\begin{array}{c}CI = \bar d - E < {\mu _d} < \bar d + E\\ = 3.67 - 5.668 < {\mu _d} < 3.67 + 5.668\\ = - 2.00 < {\mu _d} < 9.33\end{array}\)

There is 90%confidence that the limits of -2.00 and 9.33contain the mean value of the height differences.

Here clearly, one can see that 0 is included in the interval. Therefore, the value of the mean differences can also be equal to zero.

Thus, there is insufficient evidence to support the claim that differences between the heights of the presidents and their main opponents have a mean greater than 0.