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 } = 1.53 \\) (round to two decimal places as needed.) (c) determine the p - value. the p - value is (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for P - value in a right - tailed z - test

For a right - tailed z - test with test statistic \(z_0\), the P - value is \(P(Z>z_0)\), where \(Z\) is a standard normal random variable.

Step2: Use the standard normal distribution property

We know that \(P(Z > z_0)=1 - P(Z\leq z_0)\). Given \(z_0 = 1.53\).
From the standard normal table (or using a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(1.53,TRUE)), \(P(Z\leq1.53)=0.9370\).

Step3: Calculate the P - value

\(P - value=1 - 0.9370 = 0.063\)

Answer:

\(0.063\)