Question

Determine whether a normal sampling distribution can be used. If can be used, test the claim about the difference between two population proportions

p1p _ { 1 }

and

p2p _ { 2 }

at the level of significance

α\alpha

. Assume the samples are random and independent. Claim:

p1<p2;α=0.05p _ { 1 } < p _ { 2 }; \alpha = 0.05

. Sample statistics:

x1=471,n1=785x _ { 1 } = 471, n _ { 1 } = 785

and

x2=372,n2=465x _ { 2 } = 372, n _ { 2 } = 465

Solution

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Answered 4 months ago
Answered 4 months ago
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Given:

p1<p2p_1<p_2

α=0.05\alpha=0.05

x1=471x_1=471

n1=785n_1=785

x2=372x_2=372

n=2=465n=2=465

(a) The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

H0:p1p2H_0:p_1\geq p_2

Ha:p1<p2H_a:p_1<p_2

If the alternative hypothesis contains <<, then the test is left-tailed.

If the alternative hypothesis contains >>, then the test is right-tailed.

If the alternative hypothesis contains \neq, then the test is two-tailed. Left-tailed test

Determine the sample proportions. The sample proportion is the number of successes divided by the sample size:

p^1=x1n1=471785=0.6\hat{p}_1=\dfrac{x_1}{n_1}=\dfrac{471}{785}=0.6

p^2=x2n2=372465=0.8\hat{p}_2=\dfrac{x_2}{n_2}=\dfrac{372}{465}=0.8

p=x1+x2n1+n2=471+372785+465=0.6744\overline{p}=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{471+372}{785+465}=0.6744

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