Exercise 854 One in five 18yearold Americans has not graduat

Exercise 8-54

One in five 18-year-old Americans has not graduated from high school (The Wall Street Journal, April 19, 2007). A mayor of a northeastern city comments that its residents do not have the same graduation rate as the rest for the country. An analyst from the Department of Education decides to test the mayor’s claim. In particular, she draws a random sample of 80 18-year-olds in the city and finds that 20 of them have not graduated from high school. Use Table http://lectures.mhhe.com/connect/0077639472/Table/table1.jpg

a.

Compute the point estimate for the proportion of 18-year-olds who have not graduated from high school in this city. (Do not round intermediate calculations. Round your answer to 2 decimal places.)

   Point estimate   

b.

Use this point estimate to derive a 95% confidence interval for the population proportion. (Do not round intermediate calculations. Round \"z\" value to 2 decimal places and final answers to 3 decimal places.)

  Confidence interval   to    

c.

Can the mayor’s comment be justified at 95% confidence?

Yes
No

Solution

a)
No.of them have not graduated from high (x)=20
Sample Size(n)=80
Sample proportion = x/n =0.25
b)
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

Confidence Interval = [ 0.25 ±Z a/2 ( Sqrt ( 0.25*0.75) /80)]
= [ 0.25 - 1.96* Sqrt(0.002) , 0.25 + 1.96* Sqrt(0.002) ]
= [ 0.155,0.345]
c)
One in every five = 1/5 = 0.20, Yes it lies in the interval