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Sagot :
Answer:
(a) (0.9240, Â 0.956)
(b) (0.6383, 0.7017)
Step-by-step 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, [tex]\hat p[/tex] = 94%
(a) The confidence interval of a percentage is given as follows;
[tex]CI=\hat{p}\pm z\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
The z-value for 95% confidence interval = 1.96
We get;
[tex]CI=0.94\pm 1.96\times \sqrt{\dfrac{0.94 \times (1-0.94)}{846}} \approx 0.94 \pm 1.6003 \times 10^{-2}[/tex]
CI = 0.9240 ≤ [tex]\hat p[/tex] ≤ 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, [tex]\hat p[/tex] = 67% = 0.67
The 95% confidence interval is therefore;
[tex]CI=0.67\pm 1.96\times \sqrt{\dfrac{0.67 \times (1-0.67)}{846}} \approx 0.67 \pm 3.1686\times 10^{-2}[/tex]
From which we have;
CI = 0.6383 ≤ [tex]\hat p[/tex] ≤ 0.7017 = (0.6383, 0.7017)
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