Use the traditional method of hypothesis testing to test the inferences of two p

Question

Use the traditional method of hypothesis testing to test the inferences of two populations. Assume that two dependent samples have been randomly selected from normally distributed population. A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product. At the 0.05 significance level, test the claim that the recognition rates are the same in both states. a. State the null hypothesis (H_0) and the alternate (H_1): b. What is the test statistic and the critical value(s): c. Find the confidence interval values: d. Interpret the information and state what can you say about the claim?

Explanation / Answer

Solution:-

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

Null hypothesis: P1 = P2 (Claim) (Recognition rates are same in both states)
Alternative hypothesis: P1 P2 (Recognition rates are different in both states)

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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) = [(0.345878 * 558) + (0.319218 * 614)] / (558+ 614) = 389/1172 = 0.3319

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = sqrt [ 0.3319 * 0.6681 * ( 1/558 + 1/614 ) ] = 0.02754145517

z = (p1 - p2) / SE = (193/558 - 196/614)/0.02754145517 = 0.96799

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.

The z-critical values for a two-tailed test, for a significance level of = 0.05

zc = -1.96 and zc = 1.96

We use the Normal Distribution Calculator to find P(z < -0.96799), and P(z > 0.96799)

The P-Value is 0.333094.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.333094) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. We have sufficient evidence to prove the claim that the recognition rates are same in both the states.

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