Question

1. A simple random sample of 800 automobile owners in a large state revealed that 480
purchased cars from Japanese or European car companies. Construct a 95% confidence interval
for the proportion of all automobile owners in that state who purchase cars from Japanese or
European car companies.
2. There are 400 students currently studying statistics at Breakport University. The average
GPA of a random sample of 50 of these students was 3.2 with a sample standard deviation of
0.2. What is a 95% confidence interval for the average GPA of all of the statistics students?

Answer 1

1)

Sample proportion $\hat{p}$ = 480 / 800 = 0.60

95% confidence interval for p is

$\hat{p}$ - Z_$\alpha$/2 * Sqrt( $\hat{p}$ ( 1 - $\hat{p}$) / n) < p < $\hat{p}$ + Z_$\alpha$/2 * Sqrt( $\hat{p}$ ( 1 - $\hat{p}$) / n)

0.60 - 1.96 * sqrt( 0.6 * 0.4 / 800) < p < 0.60 + 1.96 * sqrt( 0.6 * 0.4 / 800)

0.5661 < p < 0.6339

95% CI is (0.5661 , 0.6339)

2)

95% confidence interval for $\mu$ is

$\bar{x}$ - t_$\alpha$/2 * S / sqrt(n) < $\mu$ < $\bar{x}$ + t_$\alpha$/2 * S / sqrt(n)

3.2 - 2.01 * 0.2 / sqrt(50) < $\mu$ < 3.2 + 2.01 * 0.2 / sqrt(50)

3.1431 < $\mu$ < 3.2569

95% CI for $\mu$ is (3.1431 , 3.2569)