Question

conduct a test at the \\( \alpha = 0.10 \\) level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the p - value. assume the samples were obtained independently from a large population using simple random sampling. test whether \\( p _ { 1 } > p _ { 2 } \\). the sample data are \\( x _ { 1 } = 126, n _ { 1 } = 257, x _ { 2 } = 135 \\), and \\( n _ { 2 } = 317 \\) (a) choose the correct null and alternative hypotheses below. a. \\( h _ { 0 } : p _ { 1 } = p _ { 2 } \\) versus \\( h _ { 1 } : p _ { 1 } > p _ { 2 } \\) b. \\( h _ { 0 } : p _ { 1 } = p _ { 2 } \\) versus \\( h _ { 1 } : p _ { 1 } < p _ { 2 } \\) c. \\( h _ { 0 } : p _ { 1 } = 0 \\) versus \\( h _ { 1 } : p _ { 1 } > p _ { 2 } \\) d. \\( h _ { 0 } : p _ { 1 } = p _ { 2 } \\) versus \\( h _ { 1 } : p _ { 1 } \
eq p _ { 2 } \\) (b) determine the test statistic. \\( z _ { 0 } = \\) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportions

The sample proportion for the first sample is $\hat{p}_{1}=\frac{x_{1}}{n_{1}}$. Substituting $x_{1} = 126$ and $n_{1}=257$, we get $\hat{p}_{1}=\frac{126}{257}\approx0.490$.
The sample proportion for the second sample is $\hat{p}_{2}=\frac{x_{2}}{n_{2}}$. Substituting $x_{2}=135$ and $n_{2} = 317$, we get $\hat{p}_{2}=\frac{135}{317}\approx0.426$.
The pooled proportion is $\hat{p}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}$. Substituting $x_{1} = 126,x_{2}=135,n_{1}=257,n_{2}=317$, we have $\hat{p}=\frac{126 + 135}{257+317}=\frac{261}{574}\approx0.455$.

Step2: Calculate the test statistic

The formula for the test statistic $z_{0}$ in a two - proportion z - test is $z_{0}=\frac{\hat{p}_{1}-\hat{p}_{2}}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$.
Substitute $\hat{p}_{1}\approx0.490,\hat{p}_{2}\approx0.426,\hat{p}\approx0.455,n_{1}=257,n_{2}=317$ into the formula:

$$
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$$

Answer:

$z_{0}\approx1.52$