You wish to test the following claim (HaHa) at a significance level of ?=0.02?=0

Question

You wish to test the following claim (HaHa) at a significance level of ?=0.02?=0.02.

      Ho:p1=p2Ho:p1=p2
      Ha:p1<p2Ha:p1<p2

You obtain a sample from the first population with 166 successes and 68 failures. You obtain a sample from the second population with 173 successes and 34 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 statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

less than (or equal to) ??

greater than ??

This test statistic leads to a decision to...

reject the null

accept the null

fail to reject the null

As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.

There is not sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.

The sample data support the claim that the first population proportion is less than the second population proportion.

There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1> P2
Alternative hypothesis: P1 < P2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.30088

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.04983
z = (p1 - p2) / SE

z = 4.28

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 4.28.

Thus, the P-value = 0.9999

Interpret results. Since the P-value (almost 1) is greater than the significance level (0.02), we have to accept the null hypothesis.

Accept the null

There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.

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