Question

The numbers of successes and the sample sizes for independent simple random samples from two populations are provided for a two-tailed test and a 95% confidence interval. Complete parts (a) through (d). Xy = 21, n = 60, X2 = 22, n2 = 100, a = 0.05 Click here to view a table of areas under the standard normal curve for negative values of Click here to view a table of areas under the standard normal curve for RoSive values of U DU X, 1 Tessan , using we provucures is not appropria D. Since ng - X, is less than 5, using the procedures is not appropriate E. Since n-x, is less than 5 using the procedures is not appropriate. c. If appropriate, use the two proportions z-test to conduct the required hypothesis test. What are the hypotheses for this test? O A. Ho: P = P2. H: P: P2 O c. Ho: P1 P2. H.Py>P2 O E. Ho: P4*P2.HPP2 OG. Using the two proportions Z-procedures is not appropriate OB. Ho: P, P2, HP, = P2 D. Ho:P P2. He:P*P2 OF. Ho: PP2. Hy: P = P2 Determine the test statistic, appropriate. Select the correct choice below and, if necessary, fill in the answer box to complete your answer. OAZ (Round to two decimal places as needed.) als OB. Using the two-proportions Z-procedures is not appropriate.

Answer 1

Answer:

c)

x1=21, n1=60, x2=22, n2=100, $\alpha$ =0.05

pl = 4 = 0.35

$\hat p2 =\frac{x2}{n2} = \frac{22}{ 100} =0.22$

x1 + x2 21 +22 nl + n260 + 100 = 0.2688

Ho: P1 = P2

Ha: P1 $\neq$ P2

formula for test statistics is

pl - p2 Vo*(1-7) *(1/n1 + 1/n2)

10.35 -0.22 (0.2688 + (1-0.2688)+(1/60 +1/100)

z = 1.796

z= 1.80

Test statistics (z) = 1.80

calculate p-value

P-Value = 2 * (1- P(z < 1.80))

calculate P(z < 1.80) using normal z table

P(z < 1.80) = 0.9641

P-Value = 2 * (1- 0.9641)

P-Value = 0.0718

since (P-Value=0.0718) > ( $\alpha$ =0.05)

Failed to reject null hypothesis.