A survey is given to students to study the cost of textbooks this semester. A random sample of 12 students had an average cost of $284.90 and a standard deviation of $96.10. Find a 95% confidence interval for the average textbook cost for all students assuming the cost of textbooks are normal.
Solution :
t /2,df = 2.201
Margin of error = E = t/2,df * (s /n)
= 2.201 * (96.10 / 12)
Margin of error = E = 61.06
The 95% confidence interval estimate of the population mean is,
- E < < + E
284.90 - 61.06 < < 284.90 + 61.06
223.84 < < 315.96
(223.84 , 315.96)
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