In: Statistics and Probability
One contributor to the cost of college education is the purchase of textbooks. Administrators at a private college are interested in estimating the average amount students spend on textbooks during the first four years at the college. A random sample of 200 students was taken showing a sample mean of X ̅ = $5,230. Suppose that past studies have indicated that the population standard deviation for the amount students spend on textbooks at this college is σ = $500.
a.Develop and interpret 90% confidence interval estimate of the population mean.
b.Develop and interpret 95% confidence interval estimate of the population mean.
c.Discuss what happens to the width of the confidence interval (margin of error) as the confidence level is increased?
Solution :
Given that,
Point estimate = sample mean = = 5230
Population standard deviation = = 500
Sample size = n = 200
(a)
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z 0.05 = 1.645
Margin of error = E = Z/2* ( /n)
= 1.645 * (500 / 200)
= 58
At 90% confidence interval estimate of the population mean is,
- E < < + E
5230 - 58 < < 5230 + 58
5172 < < 5288
(5172 , 5288 )
(b)
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = Z/2* ( /n)
= 1.96 * (500 / 200)
= 69
At 95% confidence interval estimate of the population mean is,
- E < < + E
5230 - 69 < < 5230 + 69
5161 < < 5299
(5161 , 5299)
(c)
The width of the confidence interval (margin of error) increases as the confidence level is increased .