Chapter 26, Problem 26.1AYK

(a)

To determine

To obtain the margin of error using table G and find the Tukey simultaneous $95\%$ confidence intervals for all pairwise comparisons of population means.

Answer to Problem 26.1AYK

The margin of error using table G is $85.18$ .

Explanation of Solution

In the question, it is given that a study describes a double-blind randomized experiment that assigned healthy undergraduate students to drink one of four beverages after fasting overnight: water, water and caffeine, water and glucose, water with caffeine and glucose. Also, it is given that because all four groups are the same size, the margin of error is the same for all six pairwise comparisons. Thus, the margin of error will be calculated as:

  $\begin{array}{l}{s}_p{}^2=\frac{(n-1)s_1^2+(m-1)s_2^2}{n+m-1}\\ \Rightarrow s_p=\sqrt{\frac{(18-1){78.49}^2+(18-1){76.28}^2}{18+18-1}}\\ =76.28\\ ME=m*\times s_p(\sqrt{\frac{1}{n}+\frac{1}{m}})\\ =3.350\times 76.28(\sqrt{\frac{1}{18}+\frac{1}{18}})\\ =85.18\end{array}$

Thus, the margin of error using table Gis $85.18$ . And the Tukey simultaneous $95\%$ confidence intervals for all pairwise comparisons of population means can be calculated as:

For water and water with caffeine:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(389.35-320.16)\pm 85.18\\ =(-15.99,154.37)\end{array}$

For water and water with glucose:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(389.35-318.16)\pm 85.18\\ =(-13.99,156.37)\end{array}$

For water and water with caffeine and glucose:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(389.35-336.44)\pm 85.18\\ =(-32.27,138.09)\end{array}$

For water with caffeine and water with glucose:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(320.16-318.16)\pm 85.18\\ =(-83.18,87.18)\end{array}$

For water with caffeine and water with caffeine and glucose:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(320.16-336.44)\pm 85.18\\ =(-101.46,68.9)\end{array}$

For water with glucose and water with caffeine and glucose:

  $\begin{array}{l}({\overline{x}}_i-{\overline{x}}_j)\pm m*\times s_p(\sqrt{\frac{1}{n_1}+\frac{1}{m_2}})\\ =(318.16-336.44)\pm 85.18\\ =(-103.46,66.9)\end{array}$

Hence the confidence intervals.

(b)

To determine

To explain in simple language what “ $95\%$ confidence” means for these intervals.

Explanation of Solution

In the question, it is given that a study describes a double-blind randomized experiment that assigned healthy undergraduate students to drink one of four beverages after fasting overnight: water, water and caffeine, water and glucose, water with caffeine and glucose. Also, it is given that because all four groups are the same size, the margin of error is the same for all six pairwise comparisons. Thus, in simple language what “ $95\%$ confidence” means for these intervals that we are $95\%$ confident that the reaction time of the students drinking the beverages were in between the confidence interval as calculated in part (a) for the different pairs of beverages.

(c)

To determine

To find out which pairs of means differ significantly at the overall $5\%$ significance level.

Answer to Problem 26.1AYK

None of the pairs of means differ significantly at the overall $5\%$ significance level.

Explanation of Solution

In the question, it is given that a study describes a double-blind randomized experiment that assigned healthy undergraduate students to drink one of four beverages after fasting overnight: water, water and caffeine, water and glucose, water with caffeine and glucose. Also, it is given that because all four groups are the same size, the margin of error is the same for all six pairwise comparisons. Thus, as at the $95\%$ confidence interval we have calculated the intervals for different pairs in part (a). And we can see that the zero is contained in each of the confidence intervals calculated for different pairs. Thus, for all the pairs we do not have sufficient evidence to conclude that there means differ significantly at the overall $5\%$ significance level. As the hypotheses defined in this case is as:

  $\begin{array}{l}{H}_0:{\mu }_i={\mu }_j\\ H_a:{\mu }_i\ne {\mu }_j\end{array}$

And for all the pairs we fail to reject the null hypothesis.