Question

ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. A survey of 1898 people found that 45% of the adults said that dandelions were the toughest weeds to control in their yards. Find the 95% confidence interval of the true proportion who said that dandelions were the toughest weeds to control in their yards A scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.33 feet. How large a sample is required? A large department store found that it averages 362 customers per hour. Assume that the standard deviation is 29.6 and a random sample of 40 hours was used to determine the average. Find the 99% confidence interval of the population mean.

Answer 1

1.)

z = 1.64

n = 1404

p = 323/1404 = 0.23

1 - p = 0.77

p * (1-p) = 0.177

CI = 0.23 $\pm$ 1.64 * $\sqrt{}$0.177 / 1404

CI = 0.23 $\pm$ 0.0184

CI = (0.2115, 0.2484)

2.)

Here, z = 1.96 for 95% confidence level

n = 1898

p = 0.45

1 - p = 0.55

p * (1-p) = 0.2475

CI = 0.45 $\pm$ 1.96 * $\sqrt{}$0.2475 / 1898

CI = 0.45 $\pm$ 0.0223

CI = (0.4276, 0.4723)

3.)

2 e2

e = 2

$\sigma$ = 4.33

z = 2.58

n = 6.6564 * 18.7489 / 4

n = 31.20004 = 31 ( approximately)

4.

n = 40

$\bar{x}$ = 362

$\sigma$ = 29.06

z = 2.58

Vn

Putting in above formula, we get

CI = 362 $\pm$ 2.58*29.06 / $\sqrt{}$ 40

CI = (350.145, 373.855)