Question

In a sample of 590 adults, 425 had children. Construct a 95% confidence interval for the true population proportion of adults with children.

Give your answers as decimals, to three places

Answer 1

Level of Significance,α = 0.05   

Number of Items of Interest,x =425

Sample Size,n = 590

  

Sample Proportion , p̂ = x/n = 0.7203

z -value =Zα/2 = 1.960                 [excel formula =NORMSINV(α/2)]

  

Standard Error , SE = √[p̂(1-p̂)/n] = 0.0185

margin of error , E = Z*SE = 1.960*0.0185=0.0362

  

95%Confidence Interval is

Interval Lower Limit = p̂ - E = 0.720-0.0362=0.684

Interval Upper Limit = p̂ + E =0.720+0.0362=0.757

  

95% confidence interval is (0.684< p < 0.757)