A study randomly assigned adult subjects to three exercise treatments: (1) a single long exercise period five days per week; (2) several ten-minute exercise periods five days per week; and (3) several ten-minute periods five days per week using a home treadmill. The study report contains the following summary statistics about weight loss (in kilograms) after six months of treatment:
$$\begin{array} { | l | l | l | l | }
\hline \text { Treatment } & \text { Mean } & \text { Std. dev. } & \boldsymbol { n } \
\hline \text { (1) Long periods } & 10.2 & 4.2 & 37 \
\hline \text { (2) Multiple short periods } & 9.3 & 4.5 & 36 \
\hline \text { (3) Multiple short periods with treadmill } & 10.2 & 5.2 & 42 \
\hline
\end{array}$$ The pooled standard deviation for the entire data set is 4.30 kg. The researchers want to contrast the long-period exercise regimen (1) with both short-period regimens (2 and 3). What is the 95% confidence interval for this contrast?

Question

A study randomly assigned adult subjects to three exercise treatments: (1) a single long exercise period five days per week; (2) several ten-minute exercise periods five days per week; and (3) several ten-minute periods five days per week using a home treadmill. The study report contains the following summary statistics about weight loss (in kilograms) after six months of treatment:
$$\begin{array} { | l | l | l | l | } 
\hline \text { Treatment } & \text { Mean } & \text { Std. dev. } & \boldsymbol { n } \
\hline \text { (1) Long periods } & 10.2 & 4.2 & 37 \
\hline \text { (2) Multiple short periods } & 9.3 & 4.5 & 36 \
\hline \text { (3) Multiple short periods with treadmill } & 10.2 & 5.2 & 42 \
\hline
\end{array}$$ The pooled standard deviation for the entire data set is 4.30 kg. The researchers want to contrast the long-period exercise regimen (1) with both short-period regimens (2 and 3). What is the 95% confidence interval for this contrast?

Quizplus · Accepted Answer

We are interested in contrasting the mean weight loss of the long-period exercise regimen with both short-period regimens, so we will perform a two-group contrast. 
The formula for a two-group contrast is:

$$\bar{Y}_1-\bar{Y}_2\pm(t_{\alpha/2,	ext{df}})\cdot s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

where $\bar{Y}_1$ and $\bar{Y}_2$ are the sample means, $s_p$ is the pooled standard deviation, $n_1$ and $n_2$ are the sample sizes, and $t_{\alpha/2,	ext{df}}$ is the t-value for the desired level of confidence and degrees of freedom.

We will use a 95% confidence interval, so $\alpha=0.05$ and $t_{\alpha/2,	ext{df}}=1.984$ for degrees of freedom of $df = n_1 + n_2 - 2 = 36 + 42 - 2 = 76$.

Plugging in the values from the table:

$$\begin{aligned} 
&\bar{Y}_1-\bar{Y}_2\
&=10.2-9.3\
&=0.9 
\end{aligned}$$

$$s_p= 4.30$$

$$\sqrt{\frac{1}{37}+\frac{1}{36}+\frac{1}{42}}=0.317$$

Therefore, the 95% confidence interval for the contrast is:

$$0.9\pm(1.984)(4.30)(0.317)$$

Simplifying:

$$0.9 \pm 0.45$$

Which gives us the interval:

$$(0.45, 1.35)$$

Therefore, the answer is C: 0.45 ± (1.984)(4.30) $\left( \sqrt { \frac { 1 } { 36 } + \frac { 0.25 } { 37 } + \frac { 0.25 } { 42 } } ight)$.

A Study Randomly Assigned Adult Subjects to Three Exercise Treatments