Confidence Intervals and Hypothesis Testing
Confidence interval for the difference in population means: Paired...
Jroup of medical researchers is interested in determining the effect of the drug Gravalux on a person's reaction time. To do this, the researchers select at random 13 patients who were prescribed Gravalux. Before taking Gravalux, the patients were asked to press a button as soon as a blue dot appeared on a screen. The times (in seconds) it took for the patients to press the button were recorded. Then, the patients were given Gravalux. After an hour, they were asked to repeat the blue dot experiment and their new reaction times were recorded. The data and the differences (After minus Before) are shown in the table below.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Patient & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
\hline After & 0.329 & 0.219 & 0.223 & 0.417 & 0.395 & 0.154 & 0.421 & 0.398 & 0.269 & 0.399 & 0.266 & 0.210 & 0.307 \\
\hline Before & 0.149 & 0.127 & 0.307 & 0.210 & 0.212 & 0.381 & 0.443 & 0.266 & 0.356 & 0.386 & 0.190 & 0.225 & 0.226 \\
\hline \begin{tabular}{c}
Difference \\
(After-Before)
\end{tabular} & 0.180 & 0.092 & -0.084 & 0.207 & 0.183 & -0.227 & -0.022 & 0.132 & -0.087 & 0.013 & 0.076 & -0.015 & 0.081 \\
\hline
\end{tabular}
Send data to calculator
Assume that the population of these differences in reaction times (After minus Before) is approximately normally distributed.
Construct a $95 \%$ confidence interval for $\mu_{d}$, the population mean difference in reaction times before and after taking Gravalux. Then find the lower and upper limits of the $95 \%$ confidence interval. Carry your intermediate computations to four or more decimal places. Round your answers to three or more decimal places. (If necessary, consult a list of formulas.)
Lower limit: $\square$
Upper limit: $\square$
Answer
Round the final answers to three decimal places: Lower limit: \( \boxed{-0.036} \), Upper limit: \( \boxed{0.117} \)
Step 1 ：Calculate the mean of the differences: \( \bar{d} = \frac{\sum d_i}{n} = \frac{0.18 + 0.092 - 0.084 + 0.207 + 0.183 - 0.227 - 0.022 + 0.132 - 0.087 + 0.013 + 0.076 - 0.015 + 0.081}{13} = 0.040692307692307694 \)
Step 2 ：Calculate the standard deviation of the differences: \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \)
Step 3 ：Use the t-distribution to find the critical value for a 95% confidence level with 12 degrees of freedom: \( t_{\text{critical}} = 2.1788128296634177 \)
Step 4 ：Calculate the margin of error: \( E = t_{\text{critical}} \times \frac{s_d}{\sqrt{n}} = 2.1788128296634177 \times \frac{0.12664542669423204}{\sqrt{13}} = 0.07653106540943064 \)
Step 5 ：Calculate the lower limit of the confidence interval: \( \bar{d} - E = 0.040692307692307694 - 0.07653106540943064 = -0.03583875771712294 \)
Step 6 ：Calculate the upper limit of the confidence interval: \( \bar{d} + E = 0.040692307692307694 + 0.07653106540943064 = 0.11722337310173833 \)
Step 7 ：Round the final answers to three decimal places: Lower limit: \( \boxed{-0.036} \), Upper limit: \( \boxed{0.117} \)
