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

Solution :

Given that,

Point estimate = sample mean = = 93.6

Population standard deviation =    = 4.2

Sample size = n = 34

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

Margin of error = E = Z/2 * ( /n)

Margin of error = E = 1.96 * ( 4.2 /  34 )

Margin of error = E = 1.41