Question

The National Center for Education Statistics reported that of college students work to pay for tuition and living expenses. Assume that a sample of college students was used in the study. A. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. B. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. C. What happens to the margin of error as the confidence is increased from 95% to 99%?

Answer 1

Answer:

A) = 0.0235

B) = (0.4095, 0.5305)

C) = With increase in confidence from 95% to 99%, also comes with increase in the margin of error and width of the confidence interval.

Step-by-step explanation:

Step 1:

size of random sample (n) = 450

Percentage of student working to pay tuition = 47%

Proportion of student working to pay tuition (P) = 0.47

Estimating the Standard error, $\sqrt{P(1-P)}$/$\sqrt{n}$

substituting the data in the above equation and solving will give 0.0235

Step 2:

For a 95% confidence, the confidence coefficient is 1 - ∝ = 0.95   ∴  ∝ = 0.05.

Using standard normal area table, ∝/2 = 0.05/2 = 0.025

Z (∝/2) = Z (0.025) = 1.96

95% confidence interval

P±Z(∝/2) * $\sqrt{P(1-P)} / \sqrt{n}$

= 0.47±1.96(0.025)

= 0.47±0.0461

= (0.4239, 0.5161)

Step 3:

For a 99% confidence, the confidence coefficient is 1 - ∝ = 0.99   ∴  ∝ = 0.01

Using standard normal area table, ∝/2 = 0.01/2 = 0.005

Z (∝/2) = Z (0.005) = 2.576

99% confidence interval

P±Z(∝/2) * $\sqrt{P(1-P)} / \sqrt{n}$

= 0.47±2.576(0.0235)

= 0.47±0.0605

= (0.4095, 0.5305)

Step 4:

Observe that it is on the increase.

1. at 95%, confidence intervals are (0.4239, 0.5161)

2. at 99%, confidence intervals are (0.4095, 0.5305)