Question

Section 6.1 (10 points) In January 2011, The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on a random sample of 1,018 American adults, 672 of whom agreed with the statement. [Marist Poll, “Road Rules: Re-Testing Drivers at Age 65?,” March 4, 2011.] You are asked to develop a confidence interval for the proportion of all adults nationally who think licensed drivers should be required to retake their road test once the drivers reach 65 years of age.

a) These data qualify for using a normal approximation (1-prop-Z-interval) method for developing a confidence interval for the population proportion. List the data conditions necessary for this method (e.g., how the data were collected) and show that these conditions have in fact been met.

b) Calculate a 95% confidence interval for the proportion of all adults nationally who think licensed drivers should be required to retake their road test once the drivers reach 65 years of age.

c) Using this interval, do these data provide evidence that a majority (i.e., more than 50%) of adults nationally think licensed drivers should be required to retake their road test once the drivers reach 65 years of age?

Circle one: Yes No Explain your choice of “yes” or “no.” You must reference the interval for full credit!

Answer 1

a)

Necessary conditions while constructing confidence intervals for proportions are,

1) Simple random sample: Data values are collected by simple random sampling

2) Normal Condition: sample distribution of proportion to be approximately normal,

np10

3) Independence Condition: the sample size should be 10% or less of the population (10% rule)

b)

The confidence interval for the proportion is obtained using the formula,

P1-P 95% CI pt za/21

Where,

672 = 0.66 1018 n

The critical value for z is obtained from z distribution table for significance level = 0.05

a/2=20.05/2 20.025= 1.96

0.66t 1.96/0.66(1- 0.66) 1018 95% CI

95% CI (0.631,0.689

c)

Yes, the 95% confidence interval is (63.1%,68.9%) is more than the 50%