Question

pevious attempt. Exercise 15-8 (LO15-2) The null and alternate hypotheses are: A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the s population revealed x2 to be 110. Use the 0.05 significance level to test the hypothesis. econd a. State the decisio n rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) The decision rule istojecHzis inside1.96(196 b. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 2 decimal places.) The pooled 080

Answer 1

$\hat{P_{1}} = 170 / 200 = 0.85$
$\hat{P_{2}} = 110 / 150 = 0.7333$

Test Statistic :-
$Z = ( \hat{P_{1}} - \hat{P_{2}}) / (\sqrt{\hat{P}*\hat{q (1/n1 + 1/n2)}})$
$\hat{P}$ is the pooled estimate of the proportion P
$\hat{P}$= ( x1 + x2) / ( n1 + n2)
$\hat{P}$= ( 170 + 110 ) / ( 200 + 150 )

$\hat{P}$= 0.8  

The pooled roportion

$\hat{q}= 1 - \hat{P} = 0.2$

Value of the test statistic

Z = 2.7011  $\approx$ 2.70

Test Criteria :-
Reject null hypothesis if $Z > Z_{\alpha/2}$
$Z_{\alpha} = Z_{0.05/2} = 1.96$

The decision rule is to reject H0 if z is inside 1.96 (1.96))

Outside Z < -1.96 OR Z > 1.96

$Z > Z_{\alpha/2}$ = 2.7011 > 1.96, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = 2 * P ( Z < 2.7011 )
P value = 0.007
Reject null hypothesis if P value < $\alpha = 0.05$
Since P value = 0.007 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

. Reject