In a study of 235 adults, the mean heart rate was 82 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 5 beats per minute. What is the 98% confidence interval for the mean beats per minute?

80.9 – 86.3

70.9 – 83.3

81.2 – 82.8

80.9 – 83.3

Question

In a study of 235 adults, the mean heart rate was 82 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 5 beats per minute. What is the 98% confidence interval for the mean beats per minute?

 80.9 – 86.3

 70.9 – 83.3

 81.2 – 82.8

 80.9 – 83.3

anonymous · Answer

@phi

anonymous · Answer

The $(1-\alpha)\times100\%$ confidence interval has the form
$$\left(\bar{x}-Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}},~\bar{x}+Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}}\right)$$
where $\bar{x}$ is the estimated population mean ($\bar{x}=82$), $Z_{\alpha/2}$ is the critical $z$ value for 98% confidence $\left(Z_{\alpha/2}=Z_{.02/2}=Z_{.01}\approx2.33\right)$, $\sigma$ is the sample standard deviation ($\sigma=5$), and $n$ is the sample size ($n=235$).
$$\left(82-2.33\frac{5}{\sqrt{235}},~82+2.33\frac{5}{\sqrt{235}}\right)=\left(81.24,~82.76\right)$$