A survey is being planned to determine the mean amount of time corporation
executives watch television. A pilot survey indicated that the mean time per week is
15 hours, with a standard deviation of 3.5 hours. It is desired to estimate the mean
viewing time within one-quarter hour. The 98 percent level of confidence is to be
used.
How many executives should be surveyed?
Solution:
We have σ = 3.5, E = ¼ = 0.25 and α = 1 – 0.98 = 0.02
Using Z-tables, the critical value is
Z (0.02/2) = Z (0.01) = 2.33
The sample size is given by:-
N = [Z*σ/E]²σ/E]²
= [2.33*σ/E]²3.5/0.25]²
= (32.62)²
= 1064.06
= 1065
1065 executives should be surveyed.
#2
Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for
sporting events, proms, birthdays, and other special occasions. Zack Schadek, the
owner, received a large shipment this morning. To ensure the quality of the
shipment, he selected a random sample of 300 cups. He found 15 to be defective.
(a) What is the estimated proportion defective in the population? (Round your
answer to 2 decimal places.)
Estimated proportion defective ( )
(b) What are the endpoints of a 95 percent confidence interval for the proportion