Question

To design a new advertising campaign, Ford Motor Company would like to estimate the proportion of drivers of the new Ford Fusion that are women. In a random sample of 84 Fusion owners, 48 of them were women. What is the 99% confidence interval estimating the proportion of all drivers who are women?

Options:

1)	( 0.51743 , 0.62542 )

2)	( 0.43235 , 0.71051 )

3)	( 0.44582 , 0.69704 )

4)	( 0.28949 , 0.56765 )

5)	( -0.43235 , 0.71051 )

You read an article in Golf Digest that someone with your average drive distance will have an average score of 81.91. You keep track of your scores for the next 10 rounds and see that your average score is 83.53 with a standard deviation of 2.302 strokes. You create a 95% confidence interval for your average score and it is (81.88, 85.18). Given this information, what is the best conclusion?

Options:

1)	The average score does not significantly differ from 81.91.

2)	We are 95% confident that your average score is less than 81.91.

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

4)	The percentage of rounds in which you score more than 81.91 is 95%.

5)	We are 95% confident that your average score is greater than 81.91.

Answer 1

1)

Sample proportion = 48 / 84 = 0.571

99% confidence interval for p is

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

0.571 - 2.576 * sqrt ( 0.571 * 0.429 / 84) < p < 0.571 + 2.576 * sqrt ( 0.571 * 0.429 / 84)

0.43235 < p < 0.71051

99% CI is ( 0.43235 , 0.71051 )

2)

Conclusion -

Since 81.91 contained in confidence interval, we do not have sufficient evidence to reject null hypothesis.

We conclude that the average score does not significantly differs from 81.91