Answer to Question #139570 in Statistics and Probability for Lin

Question

For a group of 100 men subjected to a stress situation, the mean number of heart beats per minute was 130, and the standard deviation was 5. Find the 90% confidence interval of the true mean.

Answer 1

Given that,

population size n = 100

population mean $\bar x$ = 130

Standard deviation $\sigma$ = 5

Since we have to calculate the Confidence Interval C.I. at 90% the value of Z$\alpha/2$ = Z_0.05 since its a 2 tailed test.

From the standard normal distribution table we get the value of Z$\alpha/2$ = Z_0.05 = 1.645

Formula for C.I = $\bar x\pm\ Z_{\alpha/2} *\frac {\sigma}{\sqrt n}\ =\ 130\ \pm\ 1.645*\frac {5}{\sqrt{100}}$

Hence we get,

90% Confidence Interval = (129.1775, 130.8225)