The following MINITAB output display presents the results of a hypothesis test on the difference between two proportions.
Test and CI for Two Proportions: P1, P2
$\begin{array} { l r r r } \text { Variable } & \mathrm { X } & \mathrm { N } & \text { Sample P } \ \text { P1 } & 43 & 107 & 0.40186916 \ \text { P2 } & 36 & 95 & 0.37894737 \end{array}$
Difference $= \mathrm { p } ( \mathrm { P } 1 ) = \mathrm { p } ( \mathrm { P } 2 )$
Estimate for differencę.02292179
$95 \%$ CI for difference:(-0.11191015, 0.15775373)
T-Test of difference $= 0 ($ vs not $= \mathbb { Z } ) = 0.33320523 \quad$ P-Value $= 0.7389795$
Can you reject $H _ { 0 }$ rejected at the $\alpha = 0.05$ level?

Question

The following MINITAB output display presents the results of a hypothesis test on the difference between two proportions.
 Test and CI for Two Proportions: P1, P2
$\begin{array} { l r r r } \text { Variable } & \mathrm { X } & \mathrm { N } & \text { Sample P } \ \text { P1 } & 43 & 107 & 0.40186916 \ \text { P2 } & 36 & 95 & 0.37894737 \end{array}$
Difference $= \mathrm { p } ( \mathrm { P } 1 ) = \mathrm { p } ( \mathrm { P } 2 )$
Estimate for differencę.02292179
$95 \%$ CI for difference:(-0.11191015, 0.15775373)
T-Test of difference $= 0 ($ vs not $= \mathbb { Z } ) = 0.33320523 \quad$ P-Value $= 0.7389795$
Can you reject $H _ { 0 }$ rejected at the $\alpha = 0.05$ level?

Quizplus · Accepted Answer

The p-value of 0.7389795 is greater than the significance level of 0.05, so we do not have sufficient evidence to reject the null hypothesis at the $\alpha = 0.05$ level.

The Following MINITAB Output Display Presents the Results of a Hypothesis