Given : In a random sample of 200 drivers 18 years and older drawn in 2010, 58 of the drivers said they texted while driving.
i.e. [tex]\hat{p}=\dfrac{58}{200}=0.29[/tex]
n= 200
Significance level : [tex]\alpha: 1-0.98=0.02[/tex]
Critical value : [tex]z_{\alpha/2}=2.33[/tex]
a) The confidence interval for population proportion is given by :-
[tex]\hat{p}\ \pm\ z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.29\pm(2.33)\sqrt{\dfrac{(0.29)(1-0.29)}{200}}\\\\\approx0.29\pm0.075\\\\=(0.29-0.075,0.29+0.075)=(0.215,\ 0.365 )[/tex]
Hence, a 98% confidence interval to estimate the actual proportion has a lower limit of 0.215 and an upper limit of 0.365.
b). The margin of error for this sample : [tex]E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
i.e. [tex](2.33)\sqrt{\dfrac{(0.29)(1-0.29)}{200}}=0.0747599662252\approx0.075[/tex]
Hence, the margin of error for this sample is 0.075.
