A group of six individuals with high blood pressure volunteered to test whether petting cats for 10 minutes can alter systolic blood pressure levels. Systolic blood pressures (in millimeters of mercury, or mmHg) were measured for each subject before and after petting cats for 10 minutes, with the following results:
$\begin{array} { c c c }
\hline \text { Individual } & \text { Before } & \text { After } \
\hline 1 & 170 & 145 \
2 & 172 & 168 \
3 & 173 & 160 \
4 & 186 & 158 \
5 & 170 & 162 \
6 & 189 & 152
\end{array}$
A researcher claims that the mean reduction in systolic blood pressure is 11 mmHg. Does the 98% confidence interval contradict this claim? (Hint: you need to find the 98% confidence interval for the mean reduction in systolic blood pressure.)

Question

A group of six individuals with high blood pressure volunteered to test whether petting cats for 10 minutes can alter systolic blood pressure levels. Systolic blood pressures (in millimeters of mercury, or mmHg) were measured for each subject before and after petting cats for 10 minutes, with the following results:
 $\begin{array} { c c c } 
\hline \text { Individual } & \text { Before } & \text { After } \
\hline 1 & 170 & 145 \
2 & 172 & 168 \
3 & 173 & 160 \
4 & 186 & 158 \
5 & 170 & 162 \
6 & 189 & 152
\end{array}$ 
A researcher claims that the mean reduction in systolic blood pressure is 11 mmHg. Does the 98% confidence interval contradict this claim? (Hint: you need to find the 98% confidence interval for the mean reduction in systolic blood pressure.)

Quizplus · Accepted Answer

To determine if the 98% confidence interval contradicts the researcher's claim of an 11 mmHg mean reduction in systolic blood pressure, we first calculate the mean reduction and the standard error, then use these to find the confidence interval.1. Calculate the mean reduction:$$ \text{Mean reduction} = \frac{(170-145) + (172-168) + (173-160) + (186-158) + (170-162) + (189-152)}{6} $$$$ \text{Mean reduction} = \frac{25 + 4 + 13 + 28 + 8 + 37}{6} $$$$ \text{Mean reduction} = \frac{115}{6} \approx 19.17 \, \text{mmHg} $$2. Calculate the standard deviation of the reductions:$$ \text{Reductions} = 25, 4, 13, 28, 8, 37 $$$$ \text{Mean} = 19.17 $$$$ \text{Variance} = \frac{(25-19.17)^2 + (4-19.17)^2 + (13-19.17)^2 + (28-19.17)^2 + (8-19.17)^2 + (37-19.17)^2}{5} $$$$ \text{Variance} \approx 158.97 $$$$ \text{Standard deviation} = \sqrt{158.97} \approx 12.61 \, \text{mmHg} $$3. Calculate the standard error:$$ \text{Standard error} = \frac{\text{Standard deviation}}{\sqrt{n}} $$$$ \text{Standard error} = \frac{12.61}{\sqrt{6}} \approx 5.15 \, \text{mmHg} $$4. Calculate the 98% confidence interval:For a 98% confidence level with 5 degrees of freedom, the t-value is approximately 2.571 (from t-distribution table).$$ \text{Margin of error} = t \times \text{Standard error} $$$$ \text{Margin of error} = 2.571 \times 5.15 \approx 13.24 \, \text{mmHg} $$$$ \text{Confidence interval} = \text{Mean reduction} \pm \text{Margin of error} $$$$ \text{Confidence interval} = 19.17 \pm 13.24 $$$$ \text{Confidence interval} \approx 5.93 \, \text{mmHg} \, \text{to} \, 32.41 \, \text{mmHg} $$Since the researcher's claim of an 11 mmHg mean reduction falls within the 98% confidence interval (5.93 mmHg to 32.41 mmHg), the confidence interval does not contradict the claim.

A Group of Six Individuals with High Blood Pressure Volunteered