Question

In: Statistics and Probability

A random group of 50 bowlers had an average score of 189. The standard deviation of...

A random group of 50 bowlers had an average score of 189. The standard deviation of all bowlers in the league was 9.1. Assume the population has a normal distribution. Find the margin of error, and then find the 95% confidence interval for the population mean

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 189

Population standard deviation = = 9.1

Sample size n =50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)
= 1.96* ( 9.1 / $\sqrt$ 50 $\sqrt$ )

Margin of error = E = 2.5224
At 95% confidence interval estimate of the population mean
is,

- E < < + E

189 - 2.5224 < <189 + 2.5224

186.4776< < 191.5224

( 186.4776, 191.5224)

orchestra answered 2 years ago

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