Find the standardized test statistic, z, to test the claim that $\mathrm { p } _ { 1 }

Question

Find the standardized test statistic, z, to test the claim that $\mathrm { p } _ { 1 } < \mathrm { p } _ { 2 }$ Assume the samples are random and independent.
Sample statistics: $\mathrm { n } _ { 1 } = 550 , \mathrm { x } _ { 1 } = 121 , \text { and } \mathrm { n } _ { 2 } = 690 , \mathrm { x } _ { 2 } = 195$
A)-2.513
B)-2.132
C)-0.985
D)1.116

Quizplus · Accepted Answer

To find the standardized test statistic $z$ for comparing two proportions, we use the formula:$$ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})}} $$where $\hat{p}_1 = \frac{x_1}{n_1}$, $\hat{p}_2 = \frac{x_2}{n_2}$, and $P = \frac{x_1 + x_2}{n_1 + n_2}$ is the pooled sample proportion.Given:- $n_1 = 550, x_1 = 121$, so $\hat{p}_1 = \frac{121}{550}$- $n_2 = 690, x_2 = 195$, so $\hat{p}_2 = \frac{195}{690}$First, calculate the sample proportions:- $\hat{p}_1 = \frac{121}{550}$- $\hat{p}_2 = \frac{195}{690}$Then, calculate the pooled sample proportion $P$:- $P = \frac{121 + 195}{550 + 690}$Finally, plug these values into the formula for $z$.After performing these calculations, you would find that the value of $z$ is approximately -2.513, which corresponds to choice A. This calculation involves using the formula for the test statistic for two proportions, which takes into account the difference between the sample proportions, the pooled proportion, and the sizes of the samples.

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

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