In: Statistics and Probability
There are 300 students enrolled in Business Statistics. Historically, exam scores are normally distributed with a standard deviation of 26.7. Your instructor randomly selected a sample of 40 examinations and finds a mean of 64.2. Determine a 90% confidence interval for the mean score for all students taking the course. 90% CI = to
Solution :
Given that,
= 64.2
= 26.7
n = 40
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Margin of error = E = Z/2* ( /n)
= 1.645 * ( 26.7/ 40)
= 6.9446
At 90% confidence interval estimate of the population mean is,
- E < < + E
64.2 - 6.9446 < < 64.2 + 6.9446
57.2554< < 71.1446
(57.2554 to 71.1446 )