Question

In: Statistics and Probability

One contributor to the cost of college education is the purchase of textbooks. Administrators at a...

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?

Solutions

Expert Solution

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 .


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