Find the weighted estimate, $\overline { \mathrm { p } }$ to test the claim that $\mathrm { p } _ { 1 } = \mathrm { p } _ { 2 } \text {. Use } \alpha = 0.05$ Assume the samples are random and independent.
Sample statistics: $\mathrm { n } _ { 1 } = 50 , \mathrm { x } _ { 1 } = 35 \text {, and } \mathrm { n } _ { 2 } = 60 , \mathrm { x } _ { 2 } = 40$
A)0.682
B)0.328
C)0.238
D)1.367

Question

Quizplus · Accepted Answer

The weighted estimate $\overline { \mathrm { p } }$ is calculated using the formula $\overline { \mathrm { p } } = \frac{x_1 + x_2}{n_1 + n_2}$, where $x_1$ and $x_2$ are the number of successes in each sample, and $n_1$ and $n_2$ are the sample sizes. Plugging in the given values, $\overline { \mathrm { p } } = \frac{35 + 40}{50 + 60} = \frac{75}{110} = 0.682$.

Find the Weighted Estimate p‾\overline { \mathrm { p } }p​

Find the Weighted Estimate $\overline { \mathrm { p } }$