Homework Soursein a survey of 645 males ages 18 to 64 396 say they going to
in a survey of 645 males ages 18 to 64, 396 say they going to the dentist in the past year construct a 90% and 95% confidence interval\'s for the population proportion
in a survey of 645 males ages 18 to 64, 396 say they going to the dentist in the past year construct a 90% and 95% confidence interval\'s for the population proportion
in a survey of 645 males ages 18 to 64, 396 say they going to the dentist in the past year construct a 90% and 95% confidence interval\'s for the population proportion
Solution
a) AT 0.90 %
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=396
Sample Size(n)=645
Sample proportion = x/n =0.614
Confidence Interval = [ 0.614 ±Z a/2 ( Sqrt ( 0.614*0.386) /645)]
= [ 0.614 - 1.64* Sqrt(0) , 0.614 + 1.64* Sqrt(0) ]
= [ 0.583,0.645]
b) AT 0.95
Confidence Interval = [ 0.614 ±Z a/2 ( Sqrt ( 0.614*0.386) /645)]
= [ 0.614 - 1.96* Sqrt(0) , 0.614 + 1.96* Sqrt(0) ]
= [ 0.576,0.652]
