Question

Steve's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. assume that the population standard deviation for his golf score is 4.2. the margin of error for a 95% confidence interval around this sample mean is ________.

Answer 1

Answer: The margin of error for a 95% confidence interval around this sample mean is 1.412.

Explanation:

We know that the formula to find the margin of error is given by :-

`E=\pm z^*{(\sigma)/(√(n))}`, where
`\sigma` = population standard deviation

n= sample size , z* = critical z-value as per confidence level.

As per given , we have

`\sigma` =4.2

n= 34

By z-table , for 95% confidence level : z* = 1.96

Then, the margin of error will be :

`E=\pm (1.96)(4.2)/(√(34))\approx\pm1.412`

Hence, the margin of error for a 95% confidence interval around this sample mean is 1.412.