## Related questions with answers

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

$p _ { 1 }$

and

$p _ { 2 }$

at the level of significance

$\alpha$

. Assume the samples are random and independent. Claim:

$p _ { 1 } = p _ { 2 }; \alpha = 0.10$

. Sample statistics:

$x _ { 1 } = 42, n _ { 1 } = 150$

and

$x _ { 2 } = 76, n _ { 2 } = 200$

Solution

Verified$H_0:p_1=p_2$

$H_a:p_1\neq p_2$

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

$\hat{p}_1=\dfrac{x_1}{n_1}=\dfrac{42}{150}=0.28$

$\hat{p}_2=\dfrac{x_2}{n_2}=\dfrac{76}{200}=0.38$

$\overline{p}=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{42+76}{150+200}=0.3371$

$\overline{q}=1-\overline{p}=1-0.3371=0.6629$

The values of $n_1\overline{p}$, $n_2\overline{p}$, $n_1\overline{q}$, $n_2\overline{q}$ have to be at least 5.

Satisfied

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