Answer: [tex](-0.2035,\ 0.1235)[/tex]
Step-by-step explanation:
The confidence interval for the difference of two population proportion is given by :-
[tex]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}}[/tex]
Given : [tex]n_1=100;\ n_2=100[/tex]
The proportion of students in the first sample replied that they turned to their mother rather than their father for help. =[tex]\dfrac{43}{100}=0.43[/tex]
The proportion of students in the second sample replied that they turned to their mother rather than their father for help. =[tex]\dfrac{47}{100}=0.47[/tex]
Significance level : [tex]\alpha=1-0.98=0.02[/tex]
Critical value : [tex]z_{\alpha/2}=2.326[/tex]
Now, the 98% confidence interval for [tex]p_1-p_2[/tex] will be :-
[tex]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)[/tex]
Hence, the 98% confidence interval for [tex]p_1-p_2[/tex] is [tex](-0.2035,\ 0.1235)[/tex]
