Question

Consider the hypothesis test below. Ho: P 1 P2 0 Ha: P 1 P20 The following results are for independent samples taken from the two populations. Sample 1 1 200 P 1 0.27 Sample 2 n 2 300 P 2 0.18 Use pooled estimator of p. a. What is the value of the test statistic (to 2 decimals)? b. What is the p-value (to 4 decimals)? Use z-table. c. With a .05, what is your hypothesis testing conclusion? Select

Answer 1

For sample 1, we have that the sample size is N 200, the number of favorable cases is X,-54, so then the sample proportion is plー#- -0.27 For sample 2, we have that the sample size is N2300, the number of favorable cases is X2-54, so then the sample proportion is p2 붊-0.18 The value of the pooled proportion is computed as p- -M-54 -0.216 Also, the given significance level is α = 0.05 The following null and alternative hypotheses need to be tested Ho:P1 P2 На:P1>P2 This corresponds to a right-tailed test, for which a z-test fortwo population proportions needs to be conducted (2) Rejection Region Based on the information provided, the significance level is α 0.05, and the critical value for a right-tailed test is z 1.64 The rejection region for this right-tailed test is R {z : z > 1.64 The z-statistic is computed as follows Pi-p 0.27-0.18 2.396 Vi(1-p)(1/m + 1/n2) 216 . (1-0.216)(1/200+ 1/300) Since it is observed that z 2.396 > zc-1.64, it is then concluded that the null hypothesis is rejected Using the P-value approach. The p-value is p 0.0083, and since p 0.00830.05, it is concluded that the null hypothesis is rejected It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion pi is greater than p2, at the 0.05 significance level. The 95% confidence interval for P1-P2 is: 0.015 < Pi-P2 < 0.165

A)

TS = 2.40

b)

p-value = 0.0083

c)

p-value < 0.05

hence we reject the null hypothesis

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