Find the standardized test statistic, z, to test the claim that $\mathrm { p } _ { 1 } \neq \mathrm { p } _ { 2 }$ Assume the samples are random and independent.
Sample statistics: $\mathrm { n } _ { 1 } = 1000 , \mathrm { x } _ { 1 } = 250 , \text { and } \mathrm { n } _ { 2 } = 1200 , \mathrm { x } _ { 2 } = 195$
A)5.09
B)2.80
C)4.76
D)3.21

Question

Quizplus · Accepted Answer

The standardized test statistic z is calculated using the formula for the difference between two proportions. Here, $\hat{p}_1 = \frac{250}{1000} = 0.25$ and $\hat{p}_2 = \frac{195}{1200} = 0.1625$. The pooled proportion $\hat{p} = \frac{250 + 195}{1000 + 1200} = 0.205$. The standard error is $\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}$. Calculating these gives a z-value of approximately 5.09.

Find the Standardized Test Statistic, Z, to Test the Claim p1≠p2\mathrm { p } _ { 1 } \neq \mathrm { p } _ { 2 }p1​=p2​

Find the Standardized Test Statistic, Z, to Test the Claim $\mathrm { p } _ { 1 } \neq \mathrm { p } _ { 2 }$