Question

SHOW ALL WORK. PLEASE HELP THIS DUE AT MIDNIGHT

Answer 1

In the first question,
Total no. of classmates(asked by Theresa for vote)=n= 100
no. of classmates who say yes= 62
proportion= p = 62/100 = 0.62
To find the 95% confidence interval use the formula,
          $( p- z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } , p+z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} })$ 
 
here,
1-α=95%
1-α=0.95
α(significance level)=0.05
α/2= 0.025
 From statistics table,
At α/2=0.025  z value is 1.96

By putting the values in above equation,
         $( 0.62- (1.96) \sqrt{ \frac{0.62(1-0.62)}{100} } ,0.62+(1.96) \sqrt{ \frac{0.62(1-0.62)}{100} }$
         =(0.5249, 0.715)

In the second question,
Total no. of highways=n= 125
no. of highways need to be repair= 75
proportion= p = 75/125 = 0.6
To find the 99% confidence interval use the formula,
          $( p- z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } , p+z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} })$ 
 
here,
1-α=99%
1-α=0.99
α(significance level)=0.01
α/2= 0.005
 From statistics table,
At α/2=0.005  z value is 2.576

By putting the values in above equation,
         $( 0.6- (2.576) \sqrt{ \frac{0.6(1-0.6)}{125} } ,0.6+(2.576) \sqrt{ \frac{0.6(1-0.6)}{125} }$
         =(0.4871, 0.7129)