Answer:
(a) (0.9240, 0.956)
(b) (0.6383, 0.7017)
Explanation:
The number of internet users involved in the survey, n = 846
The percentage of the respondents that said the internet has been good thing for them personally,
= 94%
(a) The confidence interval of a percentage is given as follows;
![CI=\hat{p}\pm z* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}](https://img.qammunity.org/2022/formulas/mathematics/high-school/imdicxwskvjcsjdvv1tn6gwc38bn7rdwg3.png)
The z-value for 95% confidence interval = 1.96
We get;
![CI=0.94\pm 1.96* \sqrt{(0.94 * (1-0.94))/(846)} \approx 0.94 \pm 1.6003 * 10^(-2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/rbrubxpuopk37dr962o5fxahwj07tmatl5.png)
CI = 0.9240 ≤
≤ 0.956 = (0.9240, 0.956)
(b) The percentage of the internet users that said the internet has directly strengthened their relationship with their families,
= 67% = 0.67
The 95% confidence interval is therefore;
![CI=0.67\pm 1.96* \sqrt{(0.67 * (1-0.67))/(846)} \approx 0.67 \pm 3.1686* 10^(-2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/fwh425upt3smun9b5dojsb5689gi6df4d7.png)
From which we have;
CI = 0.6383 ≤
≤ 0.7017 = (0.6383, 0.7017)