Question

A random sample of 100 students at a high school was asked whether they would ask their father or mother for help with a financial problem. A second sample of 100 different students was asked the same question regarding a dating problem. Forty-three students in the first sample and 47 students in the second sample replied that they turned to their mother rather than their father for help. Construct a 98% confidence interval for p1 - p2.

Answer 1

Answer:
`(-0.2035,\ 0.1235)`

Explanation:

The confidence interval for the difference of two population proportion is given by :-

`p_1-p_2\pm z_(\alpha/2)\sqrt{(p_1(1-p_1))/(n_1)+(p_2(1-p_2))/(n_2)}`

Given :
`n_1=100;\ n_2=100`

The proportion of students in the first sample replied that they turned to their mother rather than their father for help. =
`(43)/(100)=0.43`

The proportion of students in the second sample replied that they turned to their mother rather than their father for help. =
`(47)/(100)=0.47`

Significance level :
`\alpha=1-0.98=0.02`

Critical value :
`z_(\alpha/2)=2.326`

Now, the 98% confidence interval for
`p_1-p_2` will be :-

`0.43-0.47\pm(2.326)\sqrt{(0.43(1-0.43))/(100)+(0.47(1-0.47))/(100)}\\\\\approx-0.04\pm(0.1635)\\\\=(-0.04-0.1635,\ -0.04+0.1635)\\\\=(-0.2035,\ 0.1235)`

Hence, the 98% confidence interval for
`p_1-p_2` is
`(-0.2035,\ 0.1235)`