Question

7.26. An appliance manufacturer offers maintenance contracts on its major appliances. A manager wants to know what fraction of buyers of the company’s convection ovens are also buying the maintenance contract with the oven. From a random sample of 120 sales slips, 31 of the oven buyers opted for the contract.

a. The proportion of customers who buy the contract along with their oven is estimated as ____?

b. Calculate a standard error for the estimate in part a.

c. Calculate a 95% interval estimate for the true proportion of customers who buy the contract along with their oven.

d. Interpret the interval in part c.

Answer 1

Solution:

Given ,

n = 120

x = 31

a)Let Let be the sample proportion.

= x / n = 31 / 120 = 0.258

The proportion of customers who buy the contract along with their oven is estimated as 0.258

b) SE = =   

Put   in place of p

SE = =

= 0.0399

c)

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [(0.25*0.50)/n]

= 0.0783

Now the confidence interval is given by

( - E)   ( + E)

(0.258 - 0.0782)   (0.258 + 0.0782 )

0.18     0.34

Interval is (0.18 , 0.34 )

d) There is 0.95 probability that the true proportion will lie in the interval (0.18 , 0.34 )