Find the sample proportions and test statistic for equal proportions a-Dissatisf

Question

Find the sample proportions and test statistic for equal proportions a-Dissatisfied workers in two companies: x1-40, n1-100, x2-30, n2° 100, -.05, two-tailed test. 1) (Round your answers to 3 decimal places. Use Excel or your calculator or Minitab to calculate the p-value.) P2 Zcalc p-value Za/2 1.485 1382 aChoose the appropriate hypothesis. 2) HpTl-r2° 0 vs. H1-F2 * 0. Reject H0 if zcalc 1.96 Ho:r1-r2° 0 vs. H1 :-r2 #0 Reject HO if zcalc >-1.96 or zcale* 1 .96 a. b. (a- Based on the data reject H 3) True False (b-Rooms rented at least a week in advance at two hotels: x1-24. n1-200, x2 = 12, n2 = 50, = .01, left- 1) tailed test. (Round your answers to 3 decimal places. Negative values should be indicated by a minus sign. Use Excel, your calculator or Minitab to calculate the p-value.) 12 24 P2 Zcalc p-value 2.3263 Choose the appropriate hypothesis. 2) Ho: 1-r220 vs. Ho:r 1-r2 1.645 0 vs. H 1-r2 > 0. Reject H0 if zcalc

Explanation / Answer

Solution:-

c-1)

p1 = 36/480 = 0.075

p2 = 0.05

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 0.05. 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)

p = 0.062

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.0153
z = (p1 - p2) / SE

z = 1.63

zalpha = 1.96

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 1.63.

Thus, the P-value = 0.0516

Interpret results. Since the P-value (0.0516) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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