Copyright

Suppose we have independent random samples of size n_1 = 420 and n_2 = 510. The proportions of...

Question:

Suppose we have independent random samples of size {eq}n_1 = 420 {/eq} and {eq}n_2 = 510 {/eq}. The proportions of success in the two samples are {eq}p_1 = .38 {/eq} and {eq}p_2 = .43 {/eq}. Find the {eq}99 \% {/eq} confidence interval for the difference in the two population proportions.

Confidence Interval

The confidence interval is the statistical methodology to derive conclusions about the population parameter. It is determined by factors like significance level, sample size, and statistics under study. Change in value of these factors leads to a change in confidence interval limits.

Answer and Explanation:

Given Information

Sample 1

Size of Sample 1{eq}\left( {{n_1}} \right) = 420 {/eq}

Proportions of success in the sample 1 {eq}\left( {{p_1}} \right) = 0.38 {/eq}

Sample 2

Size of Sample 2{eq}\left( {{n_2}} \right) = 510 {/eq}

Proportions of success in the sample 2 {eq}\left( {{p_2}} \right) = 0.43 {/eq}


Margin of error 99% confidence interval of the difference in the two population proportions

{eq}\begin{align*} M.E &= {Z_{\dfrac{\alpha }{2}}} \times \sqrt {\dfrac{{{{\hat p}_1}\left( {1 - {{\hat p}_1}} \right)}}{{{n_1}}} + \dfrac{{{{\hat p}_2}\left( {1 - {{\hat p}_2}} \right)}}{{{n_2}}}} \\ &= {Z_{\dfrac{{0.01}}{2}}} \times \sqrt {\dfrac{{0.38\left( {1 - 0.38} \right)}}{{420}} + \dfrac{{0.43\left( {1 - 0.43} \right)}}{{510}}} \\ &= {\rm{2}}{\rm{.576}} \times \sqrt {\dfrac{{0.38\left( {1 - 0.38} \right)}}{{420}} + \dfrac{{0.43\left( {1 - 0.43} \right)}}{{510}}} \\ &= 0.0831 \end{align*} {/eq}

Therefore the 99% confidence interval of the difference of proportion is given as,

Upper limit

{eq}\begin{align*} \left( {UL} \right) &= \left( {{p_2} - {p_1}} \right) + M.E\\ &= \left( {0.43 - 0.38} \right) + 0.0831\\ &= 0.05 + 0.0831\\ &= 0.1331 \end{align*} {/eq}

Similarly lower limit

{eq}\begin{align*} \left( {LL} \right) &= \left( {{p_2} - {p_1}} \right) - M.E\\ &= \left( {0.43 - 0.38} \right) - 0.0831\\ &= 0.05 - 0.0831\\ &= - 0.0331 \end{align*} {/eq}

Hence the 99% confidence interval of difference of two proportions is 0 to 0.1331.


Learn more about this topic:

Loading...
Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
14K

Related to this Question

Explore our homework questions and answers library