Question

Is college worth it? Among a simple random sample of 300 American adults who do not have a four-year college degree and are not currently enrolled in school, 147 said they decided not to go to college because they could not afford school.

1. Calculate a 99% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. Round to 4 decimal places.

2. Suppose we wanted the margin of error for the 99% confidence level to be about 3.5%. What is the smallest sample size we could take to achieve this? Note: For consistency's sake, round your z* value to 3 decimal places before calculating the necessary sample size.

Is college worth it? Among a simple random sample of 350 American adults who do not have a four-year college degree and are not currently enrolled in school, 130 said they decided not to go to college because they could not afford school.

NOTE: While performing the calculations, do not used rounded values. For instance, when calculating a p-value from a test statistic, do not use a rounded value of the test statistic to calculate the p-value. Preserve all the decimal places at each step.

Enter at least 4 decimal places for each answer in WeBWorK.

1. A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. What are the correct hypotheses for conducting a hypothesis test to determine if these data provide strong evidence supporting this statement?

A. ?0:?=0.5H0:p=0.5, ??:?>0.5HA:p>0.5
B. ?0:?=0.5H0:p=0.5, ??:?<0.5HA:p<0.5
C. ?0:?=0.5H0:p=0.5, ??:?≠0.5HA:p≠0.5

2. Calculate the test statistic for this hypothesis test.

3. Calculate the p-value for this hypothesis test.

4. Based on the p-value, we have:
A. very strong evidence
B. little evidence
C. strong evidence
D. extremely strong evidence
E. some evidence
that the null model is not a good fit for our observed data.

Answer 1

1) = 147/300 = 0.49

At 99% confidence level, the critical value is z_0.005 = 2.575

The 99% confidence interval is

+/- z_0.005 * sqrt((1 - )/n)

= 0.49 +/- 2.575 * sqrt(0.49(1 - 0.49)/300)

= 0.49 +/- 0.0743

= 0.4157, 0.5643

2) Margin of error = 0.035

or, z_0.005 * sqrt(p(1 - p)/n) = 0.035

or, 2.575 * sqrt(0.49(1 - 0.49)/n) = 0.035

or, n = (2.575 * sqrt(0.49 * 0.51)/0.035)^2

or, n = 1353

1) Option - B) H₀: P = 0.5

                      H_A: P < 0.5

2) = 130/350 = 0.3714

The test statistic is

3) P-value = P(Z < -4.8118)

                 = 0.0000

4) Option - D) extremely strong evidence