Question

Refer to the following scenario:You want to see if there is a difference between the exercise habits of Science majors and Math majors. You survey 135 science majors, and find out that 82 of them regularly exercise. You survey 92 math majors, and find out that 41 of them regularly exercise. You test your hypothesis that the proportions are different at the 1% significance level.1. Which of the following is the correct null hypothesis? A. H0 : A = 0 B. H0 : p = 0 C. H0: P1 = P2 D. H0 : H1 = 12 2. Which of the following is the correct alternative hypothesis? A. H0.: P1 + P2 B. H0 : P1 > P2 C. H0 : Pi + P2 D. H0 : M1 is not equal to M2 3. What is the pooled proportion of Science and Math majors that regularly exercise? 4. What is the p-value of your test? 5. State the conclusion of your test in context?6. What is a 99% confidence interval for the difference in the true proportions of Science and Math majors who regularly exercise?

Answer 1

Answer:

1. H0: P1 = P2

2. Ha: P1 ≠ P2

3. pooled proportion p = 0.542

4. P-value = 0.0171

5. The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

6. The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

Step-by-step explanation:

We should perform a hypothesis test on the difference of proportions.

As we want to test if there is significant difference, the hypothesis are:

Null hypothesis: there is no significant difference between the proportions (p1-p2 = 0).

Alternative hypothesis: there is significant difference between the proportions (p1-p2 ≠ 0).

The sample 1 (science), of size n1=135 has a proportion of p1=0.607.

$p_1=X_1/n_1=82/135=0.607$

The sample 2 (math), of size n2=92 has a proportion of p2=0.446.

$p_2=X_2/n_2=41/92=0.446$

The difference between proportions is (p1-p2)=0.162.

$p_d=p_1-p_2=0.607-0.446=0.162$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{82+41}{135+92}=\dfrac{123}{227}=0.542$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.542*0.458}{135}+\dfrac{0.542*0.458}{92}}\\\\\\s_{p1-p2}=\sqrt{0.001839+0.002698}=\sqrt{0.004537}=0.067$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.162-0}{0.067}=\dfrac{0.162}{0.067}=2.4014$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=2\cdot P(z>2.4014)=0.0171$

As the P-value (0.0171) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

We want to calculate the bounds of a 99% confidence interval of the difference between proportions.

For a 99% CI, the critical value for z is z=2.576.

The margin of error is:

$MOE=z \cdot s_{p1-p2}=2.576\cdot 0.067=0.1735$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.162-0.1735=-0.012\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.162+0.1735=0.335$

The 99% confidence interval for the difference between proportions is (-0.012, 0.335).