Question

10. College Score Card Data A reporter has claimed that the majority of Colleges in the United States are Private For-Profit colleges. Using the College Score Card data as a sample and a 0.05 level of significance, test the claim that the majority of Colleges in the USA are private for-profit. a) Use the College Score Card data to find for the proportion of colleges in the sample which are private for-profit. Þ: 5339 (Round 4 decimal places) b) Complete the following areas. Label the critical value and test statistic on the curve and shade the rejection region(s). Ho: H: Type of Test: Test Statistic: (2 decimal places) Critical Value(s): (2 decimal places) P-Value: (4 decimal places) Reject or Fail to Reject: Conclusion (express in a sentence in the context of the problem

Answer 1

10)

a)

the sample proportion p = 0.5339

b)

Answer:

$H_0:p=0.50$

H:p > 0.50

Type of Test: z test

Test statistic: z = 3.19

Critical value: z = 1.64

P-value: p-value = 0.0007

Reject or Fail to Reject: Reject the null hypothesis

Explanation:

Hypothesis:

The null and alternative hypotheses are defined as,

$H_0:p=0.50$

H:p > 0.50

This is a right-tailed test

Type of Test:

To test whether there is a significant difference in results from the survey, the z test for the one proportion is used since the sample data values satisfy the normality condition $np\geq10$

Test statistic:

The z-statistic is obtained using the formula,

$z=\frac{p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

0.5339 -0.50 10.50(1-0.50) 2212

$z=3.19$

Critical value:

Let the significance level = 0.05

The z critical value is obtained from the standard normal distribution table for significance level = 0.05 for one-tailed distribution.

$z_c=z_{1-\alpah}=z_{1-0.05}=z_{0.95}=1.64$

P-value:

The p-value is obtained from z-distribution table for z = 3.19 for right-tailed test

$\text{P-value}=0.0007$

Reject or Fail to Reject:

Since the P-value is 0.0007 is less than 0.05 at 5% significant level, the null hypothesis is rejected.

12)

a)

Answer:

$\widehat{p}=0.203$

Explanation:

$\widehat{p}=\frac{205}{1008}=0.203$

b)

Answer:

$H_0:p=0.50$

H:p > 0.50

Type of Test: z test

Test statistic: z = 0.27

Critical value: z = 1.64

P-value: p-value = 0.3945

Reject or Fail to Reject: Fail to Reject the null hypothesis

Explanation:

Hypothesis:

The null and alternative hypotheses are defined as,

$H_0:p=0.20$

$H_1:p>0.20$

This is a right-tailed test

Type of Test:

To test whether there is a significant difference in results from the survey, the z test for the one proportion is used since the sample data values satisfy the normality condition $np\geq10$

Test statistic:

The z-statistic is obtained using the formula,

$z=\frac{p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

$z=\frac{0.2034-0.20}{\sqrt{\frac{0.20(1-0.20)}{1008}}}$

$z=0.268$

Critical value:

Let the significance level = 0.05

The z critical value is obtained from the standard normal distribution table for significance level = 0.05 for one-tailed distribution.

$z_c=z_{1-\alpah}=z_{1-0.05}=z_{0.95}=1.64$

P-value:

The p-value is obtained from z-distribution table for z = 0.268 for right-tailed test

$\text{P-value}=0.3945$

Reject or Fail to Reject:

Since the P-value is 0.3945 is greater than 0.05 at 5% significant level, the null hypothesis is not rejected.