Answer: (0.881, 0.919)
Step-by-step explanation:
Let p be the population proportion of respondents who say the Internet has been a good thing for them personally.
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where z* = Critical value ,
[tex]\hat{p}[/tex] = Sample proportion
n = sample size.
As per given , we have
n= 957
[tex]\hat{p}=0.90[/tex]
By z-table , Critical value for 95% confidence interval : z*= 1.96
Then, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. will be :
[tex]0.90\pm 1.96\sqrt{\dfrac{0.90(1-0.90)}{957}}[/tex]
[tex]0.90\pm 1.96\sqrt{0.0000940438871473}[/tex]
[tex]0.90\pm 1.96(0.00969762275753)[/tex]
[tex]\approx0.90\pm 0.019\\\\=(0.90-0.019,\ 0.90+0.019)[/tex]
[tex](0.881,\ 0.919)[/tex]
Hence, the required 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally = (0.881, 0.919)