Question

Assume the following information X1 -55 n] = 100 X2-80 n2-200 We wish to test if the first population proportion is more than the second one at - 0.05 significant level Answer from quetion 1-5 The pooled estimator for overall proportion Pe 0.450 0.250 0.35.CO 0.52.00

Answer 1

Calculation:

First Population:

X1 55 pl = = nl - 100 = 0.55

ql = 1-pl = 1-0.55 = 0.45

Second Population:

= = 0.40

Pooled sample Proportion:

Pa (nl* pl + n2 * p2) nl + n2

P= (100 * 0.55 + 200 * 0.40) 100 + 200

P = 0.45

The Pooled estimate for overall proportion is: P = 0.45

ANSWER: A. 0.45

Hypothesis:

Ho: P1 = P2

Ha: P1 > P2

Test statistic:

Z =- pl - p2 PQ (1+) 0.55-0.40 10.45 * 0.55 (Too + 2) ==2.46

Critical value:

Z $\alpha$ = Z $\alpha$ = 1.645 .........Using standard Normal table

Conclusion:

Test statistic (Z) > Critical value, 2.46 > 1.465, That is Reject Ho at 5% level of significance.

Therefore, There is Suffcient evidence to conclude that, the First population proportion is more than the second one at 0.05 significance level.