Question

3) A TV news reporter says that a proposed constitutional amendment is likely to win approval in the upcoming election because in a poll of 1500 likely voters, 783 said they would vote in favor of the constitutional amendment.

a) Calculate a 98% confidence interval for the percentage of voters that will vote in favor of the constitutional amendment.

b) What is the marginal error for a 98% confidence interval?

c) Does the poll support the news reporters’ claim that the amendment is likely to pass (ie. does the interval appear to agree with a vote over 50%)? Why or why not?

Answer 1

Solution:

n = 1500

x = 783

= 783/1500 = 0.522

a)

c = 98% = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.01

= 2.326 (use z table)

Now , the margin of error is given by

E = /2 *  

= 2.326* [0.522 *(1 - 0.522)/1500]

= 0.02999944917

Now the confidence interval is given by

( - E)   ( + E)

(0.522 - 0.02999944917)   (0.522 + 0.02999944917)

0.4920 0.5520

Required 98% Confidence Interval is (49.20 % , 55.20 %)

b)

Margin of error =

E = /2 *  

= 2.326* [0.522 *(1 - 0.522)/1500]

= 0.0300

Margin of error = 0.0300

c)

claim : p > 50%

Observe the interval (49.20 % , 55.20 %)

50% is in the interval ,. and lower limit is less than 50%

So ,

The interval does not appear to agree with a vote over 50%.