Answer:
Step-by-step explanation:
Given that a research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=1096 and x=542 who said yes.
i.e. sample size = [tex]n=1096\\[/tex]
who said yes =[tex]x=542\\[/tex]
Sample proportion = [tex]p=\frac{542}{1096} =0.4945[/tex]
Std error of proportion = [tex]\sqrt{\frac{p(1-p)}{n} } \\=\sqrt{\frac{0.4945(1-0.4945)}{1096} } \\=0.0151[/tex]
Margin of error 95% [tex]= 1.96 (se)\\=0.0296[/tex]
Hence confidence interval for proportions
=[tex](0.4945-0.0296, 0.4945+0.0296)\\= (0.4649,0.5241)[/tex]
a) Point estimate [tex]=0.4945[/tex]
B) Margin of error = [tex]0.0296[/tex]
C) Conf interval = [tex](0.4649,0.5241[/tex]
D) We are 95% confident that for large random samples representing the population, the proportion of yes will lie between these two values.