Find the requested confidence interval. A random sample of 30 long distance runners aged 20-25 was...

Find the requested confidence interval. A random sample of 30 long distance runners aged 20-25 was selected from a running club. The mean of resting heart rates (in beats per minute) of the runners taken in this sample is 68.07 Estimate the mean resting heart rate for the population of long distance runners aged 20-25. Assuming ? = 8.03, give the 95.44% confidence interval for the population mean.

Solutions

Expert Solution

Solution :

Given that,

= 68.07

= 8.03

n = 30

At 95.44% confidence level the z is ,

= 1 - 95.44% = 1 - 0.9544= 0.0456

/ 2 = 0.0456 / 2 = 0.0228

Z_/2 = Z_0.0228 = 1.999

Margin of error = E = Z_/2* ( $\sqrt$ /n)

= 1.999 * (8.03 / $\sqrt$ 30 ) = 2.93

At 95.44% confidence interval estimate of the population mean is,

- E < < + E

68.07 - 2.93 < < 68.07 + 2.93

65.14 < < 71.00

(65.14 , 71.00)

orchestra answered 1 year ago

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