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,
[tex]( p- z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } , p+z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} }) [/tex]
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,
[tex]( 0.62- (1.96) \sqrt{ \frac{0.62(1-0.62)}{100} } ,0.62+(1.96) \sqrt{ \frac{0.62(1-0.62)}{100} } [/tex]
=(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,
[tex]( p- z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} } , p+z_{ \alpha /2} \sqrt{ \frac{p(1-p)}{n} }) [/tex]
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,
[tex]( 0.6- (2.576) \sqrt{ \frac{0.6(1-0.6)}{125} } ,0.6+(2.576) \sqrt{ \frac{0.6(1-0.6)}{125} } [/tex]
=(0.4871, 0.7129)
