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}
= Z_{0.025} = 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 .

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