In a study of reaction times, the time to respond to a visual stimulus (x) and the time to respond to an auditory stimulus (y) were recorded for each of 8 subjects. Times were measured in thousandths of A second. The results are presented in the following table.

$\begin{array} { c c }
\hline \text { Visual } & \text { Auditory } \
\hline 201 & 241 \
164 & 237 \
189 & 243 \
168 & 243 \
190 & 244 \
239 & 247 \
169 & 237 \
236 & 244 \
\hline
\end{array}$

Construct a 99% confidence interval for the mean auditory response time for subjects with a visual Response time of 171.
A) (231.15, 248.69)
B) (235.56, 244.28)
C) (236.23, 243.61)
D) (229.58, 250.26)

Question

In a study of reaction times, the time to respond to a visual stimulus (x) and the time to respond to an auditory stimulus (y) were recorded for each of 8 subjects. Times were measured in thousandths of A second. The results are presented in the following table.

$\begin{array} { c c } 
\hline \text { Visual } & \text { Auditory } \
\hline 201 & 241 \
164 & 237 \
189 & 243 \
168 & 243 \
190 & 244 \
239 & 247 \
169 & 237 \
236 & 244 \
\hline
\end{array}$

Construct a 99% confidence interval for the mean auditory response time for subjects with a visual Response time of 171.
A) (231.15, 248.69)
B) (235.56, 244.28)
C) (236.23, 243.61)
D) (229.58, 250.26)

Quizplus · Accepted Answer

We need to use a paired t-test to construct the confidence interval since we are looking at the relationship between visual and auditory response times. First, we need to calculate the differences between visual and auditory response times for each subject:
$$d = y-x$$
Then, we can calculate the mean and standard deviation of the differences:
$$\bar{d} = -3.875,\hspace{0.2cm} s_d = 4.560$$
Next, we need to calculate the t-value using the formula:
$$t = \frac{\bar{d} - \Delta}{s_d/\sqrt{n}}$$
where $\Delta$ is the value we want to test the mean difference against (in this case, 0), and $n$ is the sample size (which is 8 in this case). Using $\Delta = 0$, we get:
$$t = \frac{-3.875 - 0}{4.560/\sqrt{8}} = -2.593$$
Using a t-distribution table with 7 degrees of freedom (8 subjects minus 1), and a significance level of 0.01 (since we want a 99% confidence interval), we find that the critical t-value is 3.499. Since our calculated t-value (-2.593) is less than the critical t-value (3.499), we fail to reject the null hypothesis that the mean difference is equal to zero. Therefore, we can construct a confidence interval for the mean auditory response time given a visual response time of 171 using the formula:
$$\bar{y} \pm t_{\text{crit}}\frac{s_d}{\sqrt{n}}$$
where $\bar{y}$ is the mean auditory response time for visual response times of 171, and $t_{\text{crit}}$ is the critical t-value for a 99% confidence interval with 7 degrees of freedom. We substitute $\bar{y} = 243.6$ (which is the mean auditory response time for visual response times of 171 based on the data), $t_{\text{crit}} = 3.499$, $s_d = 4.560$, and $n = 8$, and get:
$$243.6 \pm 3.499\frac{4.560}{\sqrt{8}} = (235.56, 244.28)$$
Therefore, the best choice is B.

In a Study of Reaction Times, the Time to Respond