Question

Question 33 3 pts Difference of two proportions. A study was done to determine computer usage of workers by age using data from the General Social Survey. The survey asked respondents if they used a computer at work. The researchers did not have a strong prior belief that one group would be higher than another. On the one hand, older workers potentially had more experience and their use might be higher. On the other, it might be reasonable to say that younger workers had more experience in using computers in school. The decision was to test if the two groups are different from each other. To test this, they broke the sample of people working full-time into two age groups. The first group was aged 25 to 40 and the second group was aged 41 to 65. The data are given below. Use alpha 05 for your test. 25 to 40 41 to 65 Number answering Yes 384 296 400 500 Since the Null Value is the difference between the proportions for each group is zero, we will pool the variance. What is the pooled Variance for this problem. I want just a numerical answer. Use 4 decimal places for your answer, and use the proper rules of rounding.

Answer 1

solution:

given that

We will set up null hypothesis

$H_{0}:p_{1}=p_{2}$

$H_{0}:p_{1}\neq p_{2}$

The first group aged 25 to 40 answering "YES" is equal to 296 out of 400

$\therefore p_{1}=\frac{X1}{N1}=\frac{296}{400}=0.74$

The second group aged 41 to 65 answering "YES" is equal to 384 out of 500

$\therefore p_{2}=\frac{X2}{N2}=\frac{384}{500}=0.768$

Under the null hypothesis the test statistics is

$Z=\frac{p_{1}-p_{2}}{\sqrt{P(1-P)\left ( \frac{1}{n1} + \frac{1}{n2} \right )}}\sim N(0,1)$

Where   $P=\frac{X1+X2}{N1+N2}=\frac{296+384}{400+500}=\frac{680}{900}=0.756$

There fore pooled variance is obtained as follow

$\sigma ^2= P(1-P)\left ( \frac{1}{n1} + \frac{1}{n2} \right )$

$\sigma ^2= 0.756(1-0.756)\left ( \frac{1}{400} + \frac{1}{500} \right )$

$\sigma ^2= 0.00079947$

$\therefore Z=\frac{0.74-0.768}{\sqrt{0.756(1-0.756)\left ( \frac{1}{400} + \frac{1}{500} \right )}}$

$Z=-0.97$

and its corresponding $P-Value = 0.3490$

Since calculated $P-Value = 0.3490$ is greater then calculated 0.05 hence we will accept the null hypothesis and conclude that P1 = P2.

thank you