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IgorC [24]
3 years ago
13

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?
Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0

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).

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