You wish to test the following claim (H) at a sipnificance level of a 0.10 1:27:

Question

You wish to test the following claim (H) at a sipnificance level of a 0.10 1:27:37 remai You obtain a sample from the first population with 161 successes and 400 failures. You obtain a sample from the second population with 213 successes and 417 failures. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution What is the test statistic for this sample? (Report answer accurate to three decimal places test statisfic What is the p-value for this sample? Report answer accurate to four decimal places) p-value The p-value i less than (or equal to)a greater than a This test statistic leads to a decision to.. reject the nu accept the u fail to reject the null

Explanation / Answer

Solution:

Here, we have to use z test for two population proportions. The null and alternative hypotheses are given as below:

H0: p1 = p2

Ha: p1 < p2

We are given ? = 0.10

X1 = 161, N = 161+400 = 561

X2 = 213, N = 213+417 = 630

Sample proportions are given as below:

P1 = X1/N1 = 161/561 = 0.286987522

P2 = X2/N2 = 213/630 = 0.338095238

P1 – P2 = 0.286987522 - 0.338095238 = -0.05110772

Test statistic formula is given as below:

Z = (P1 – P2) / sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = -0.05110772 / sqrt[(0.286987522*(1 - 0.286987522)/561) + (0.338095238*(1 - 0.338095238)/630)]

Z = -0.05110772 / sqrt((0.286987522*(1 - 0.286987522)/561) + (0.338095238*(1 - 0.338095238)/630))

Z = -1.90471

Test statistic = -1.905

P-value = 0.0287

(By using z-table)

? = 0.10

P-value < ? = 0.10

The P-value is less than (or equal to) ?.

The test statistic leads to a decision to reject the null hypothesis.

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