41) A study revealed that the 30-day readmission rate was 33.7 percent for 358 p

Question

41)

A study revealed that the 30-day readmission rate was 33.7 percent for 358 patients who received after-hospital care instructions (e.g., how to take their medications) compared to a readmission rate of 42.0 percent for 358 patients who did not receive such information.

Find the p-value for the test. (Do not round the intermediate calculations and round x1 and x2 to the nearest whole numbers. Round your answer to 4 decimal places.)

41)

A study revealed that the 30-day readmission rate was 33.7 percent for 358 patients who received after-hospital care instructions (e.g., how to take their medications) compared to a readmission rate of 42.0 percent for 358 patients who did not receive such information.

Find the p-value for the test. (Do not round the intermediate calculations and round x1 and x2 to the nearest whole numbers. Round your answer to 4 decimal places.)

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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 0.05. 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).

p1 = 33.7% = 0.337, n1 = 358

p2 = 42% = 0.42, n2 = 358

p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.337 * 358) + (0.42 * 358)] / (358 + 358) = 0.3785

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = sqrt [ 0.3785 * 0.6215 * ( 1/358 + 1/358 ) ] = 0.0362515844385

z = (p1 - p2) / SE = (0.337 - 0.42)/0.0362515844385 = -2.28955509906 or -2.29

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 -2.29 or greater than 2.29.

We use the Normal Distribution Calculator to find P(z < -2.29)

The P-Value is 0.022079 or 0.0221

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