The confidence interval for population proportion (p) is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] = sample proportion
n= Sample size
z*= Critical z-value.
Let p = population proportion of teens that would report texting with their friends every day.
As per given , we have
n=799
[tex]\hat{p}=\dfrac{392}{799}=0.49[/tex]
Critical z value for 95% confidence = z* =1.96
Now , the 95% confidence interval for the population proportion p that would report texting with their friends every day. would be:
[tex]0.49\pm 1.96\sqrt{\dfrac{0.49(1-0.49)}{799}}[/tex]
[tex]0.49\pm 1.96\sqrt{0.000312765957447}[/tex]
[tex]0.49\pm 1.96(0.01768519)[/tex]
[tex]0.49\pm 0.03466 =(0.49-0.03466,\ 0.49+0.03466)[/tex]
[tex]=(0.45534,\ 0.52466)\approx(0.46,\ 0.52)[/tex]
95% confidence interval for the population proportion p: (0.46, 0.52)
Interpretation: We are 95% confident that the true population proportion p of teens that would report texting with their friends every day lies in (0.46, 0.52).
