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: $(-0.2035,\ 0.1235)$

Step-by-step explanation:

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

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

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. =$\dfrac{43}{100}=0.43$

The proportion of students in the second sample replied that they turned to their mother rather than their father for help. =$\dfrac{47}{100}=0.47$

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 :-

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

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