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 ________.
Accepted Solution
A:
Answer: The margin of error for a 95% confidence interval around this sample mean is 1.412.Step-by-step explanation:We know that the formula to find the margin of error is given by :-[tex]E=\pm z^*{\dfrac{\sigma}{\sqrt{n}}}[/tex], where [tex]\sigma[/tex] = population standard deviationn= sample size , z* = critical z-value as per confidence level.As per given , we have[tex]\sigma[/tex] =4.2n= 34By z-table , for 95% confidence level : z* = 1.96Then, the margin of error will be :[tex]E=\pm (1.96)\dfrac{4.2}{\sqrt{34}}\approx\pm1.412[/tex]Hence, the margin of error for a 95% confidence interval around this sample mean is 1.412.