Question

The branch manager of a nationwide bookstore chain (located near a college campus) wants to study characteristics of her store’s customers. She decides to focus on two variables: the amount of money spent by customers (on items other than textbooks) and whether the customers would consider purchasing educational DVDs related to graduate preparation exams such as the GMAT, CRE, or LSAT. The results from a sample of 70 customers are as follows: Amount spent; mean = $28.52, standard deviation = $11.39; 28 customers stated that they would consider purchasing the educational DVDs.

a. Construct a 95% confidence interval estimate for the population mean amount spent in the bookstore

b. Construct a 90% confidence interval estimate for the population proportion of customers who would consider purchasing educational DVDs

Assume the branch manager wants to of another store in the chain (also located near a college campus) wants to conduct a similar survey in his store. Answer the following questions:

c. What sample size is needed to have 95% confidence of estimating the population mean amount spent in this store to within plus/minus $2 if the standard deviation is assumed to be $10?

d. How many customers need to be selected to have 90% confidence of estimating the population proportion who would consider purchasing the educational DVDs to within plus/minus 0.04

Answer 1

a).

Standard error of mean = S / = 11.39/ = 1.3614

Degree of freedom = n-1 = 70-1 = 69

Critical value of t at 95% confidence interval and df = 69 is 1.99

Margin of error = t * Std error = 1.99 * 1.3614= 2.7092

95% confidence interval of mean amount spent in the pet supply store

(28.52- 2.7092, 28.52+ 2.7092)

(25.8108, 31.2292)

b).

Sample proportion, p = 28/70 = 0.4

Standard error of sample proportion, SE = = 0.0585

Z value for 90% confidence interval is 1.645

Margin of error = Z * SE = 1.645 * 0.0585= 0.0962

90% confidence interval is ,

(0.4- 0.0962, 0.4+ 0.0962)

(0.3038, 0.4962)

c).

Margin of error, E = 2

Z value for 95% confidence interval is 1.96

Sample size, n = (z / E)2  = (1.96 * 10 / 2)2

= 96

d).

Margin of error, E = 0.04

Z value for 90% confidence interval is 1.645

Sample size, n = p(1-p) (z/ E)2  = 0.4* (1-0.4) * (1.645 / 0.04)2

= 406

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