An economist wishes to test whether there is any difference in the proportion of

Question

An economist wishes to test whether there is any difference in the proportion of households receiving Medicare for white households with less than $5,000 income per year and white households with $15,000 or more income per year. A random sample from households with less than $5,000 income per year is taken, and a random sample of households with $15,000 or more income is taken. Use a 0.01 significance level to determine if there is sufficient evidence to conclude that a difference in proportions exists. Calculate the standard error of estimate.

The results of the samples are:

Less than $5,000                          $15,000 or more

Sample size 200                                                  150

Medicare Recipients                      157                                                  139

a. Use a 0.01 significance level to determine if there is sufficient evidence to conclude that a difference in proportions exists. What is the random variable?

b. What are the null and alternative hypotheses?

c. What test will you use? What is the value of the test statistic?

d. What is the decision rule?

e. What is the conclusion?

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 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 0.01. 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 = 0.8457

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

SE = 0.03902

z = (p1 - p2) / SE

z = - 3.63

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 two-tailed test, the P-value is the probability that the z-score is less than - 3.63 or greater than 3.63.

Thus, the P-value = 0.0002

Interpret results. Since the P-value (0.0002) is less than the significance level (0.01), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is sufficient evidence to conclude that a difference in proportions exists.

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