The average of a random sample of 10 scores on a college placement exam is x ¯= 75, and the sample standard deviation is s = 8.4. Assuming that the collection of all scores on the exam is approximately normally distributed, find a 90 percent confidence interval for the mean score.
Solution :
Given that,
t /2,df = 1.833
Margin of error = E = t/2,df * (s /n)
= 1.833 * (8.4 / 10)
Margin of error = E = 4.9
The 90% confidence interval estimate of the population mean is,
- E < < + E
75 - 4.9 < < 75 + 4.9
70.1 < < 79.9
(70.1 , 79.9)
Get Answers For Free
Most questions answered within 1 hours.