Question

A U.S. census bureau pollster noted that in 462 random households surveyed, 267 occupants owned their own home. What is the 99% confidence interval estimate of the proportion of American households who own their own home?

Question 9 options:

1)	( 0.52447 , 0.63138 )

2)	( 0.55494 , 0.6009 )

3)	( 0.51873 , 0.63711 )

4)	( -0.51873 , 0.63711 )

5)	( 0.36289 , 0.48127 )

The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 64.08. In a random sample of 22 golfers that visited his course, the sample mean was 48.69 and the standard deviation was 8.026. Using this information, the owner calculated the confidence interval of (43.85, 53.53) with a confidence level of 99%. Which of the following statements is the best conclusion?

Question 10 options:

1)	The average age of all golfers does not significantly differ from 64.08.

2)	We are 99% confident that the average age of all golfers that play on the golf course is greater than 64.08

3)	The percentage of golfers with an age greater than 64.08 is 99%.

4)	We are 99% confident that the average age of all golfers that play on the golf course is less than 64.08

5)	We cannot determine the proper interpretation based on the information given.

Answer 1

1)

Sample proportion = 267 / 462 = 0.57792

99% Confidence interval for p is

- Z * sqrt ( ( 1 - ) / n) < p < + Z * sqrt ( ( 1 - ) / n)

0.57792 - 2.576 * sqrt( 0.57792 * 0.42208 / 462) < p < 0.57792 + 2.576 * sqrt( 0.57792 * 0.42208 / 462)

0.51873 < p < 0.63711

99% CI is ( 0.51873 , 0.63711 )

2)

Since all values in confidence interval less than 64.08,

We conclude that,

we are 99% confident that the average age of all golfers that play on a golf course is less than 64.08