Find the weighted estimate, $\overline { \mathrm { p } }$ tto test the claim that $\mathrm { p } _ { 1 } \neq \mathrm { p } _ { 2 } \text {. Use } \alpha = 0.02$ 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)0.202
B)0.789
C)0.110
D)0.138

Question

Quizplus · Accepted Answer

The weighted estimate, $\overline { \mathrm { p } }$, is calculated using the formula $\overline { \mathrm { p } } = \frac { x _ { 1 } + x _ { 2 } } { n _ { 1 } + n _ { 2 } }$. Plugging in the given values, we get $\overline { \mathrm { p } } = \frac { 250 + 195 } { 1000 + 1200 } = \frac { 445 } { 2200 } = 0.202$.

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

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