Question

1. A questionnaire about study habits was given to a random sample of students taking a
large introductory statistics class. Students were asked if they studied most nights in a week: yes or
no. The results are that the sample proportion of students who answered “yes” (?̂) was 0.4.

Consider the same scenario as in Problem 1.
a) Calculate a confidence interval for sample size n = 1000, ?̂= 0.4, and the following confidence
levels: 80%, 90%, and 99%. You can use your confidence interval from Problem 1 with n = 1000
for 95% so you don’t have to recalculate it.
b) Answer the question: As the confidence level increases from 80% to 99% what happens to the
width of the confidence level?

Answer 1

(b) = 0.4, n = 1000

The Confidence Interval is given by ME, where

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(i) The 80% Confidence Interval

The Zcritical (2 tail) for = 0.2, is 1.281

The Lower Limit = 0.4 - 0.018 = 0.382

The Upper Limit = 0.4 + 0.018 = 0.418

The 80% Confidence Interval is (0.382 , 0.418)

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(ii) The 90% Confidence Interval

The Zcritical (2 tail) for = 0.10, is 1.645

The Lower Limit = 0.4 - 0.023 = 0.377

The Upper Limit = 0.4 + 0.023 = 0.423

The 90% Confidence Interval is (0.77 , 0.423)

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(iii) The 99% Confidence Interval

The Zcritical (2 tail) for = 0.01, is 2.576

The Lower Limit = 0.4 - 0.036 = 0.364

The Upper Limit = 0.4 + 0.036 = 0.436

The 99% Confidence Interval is (0.364 , 0.436)

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(b) The Width of the confidence interval is dependent on the value of the Margin of Error (ME). As ME increases the width increase and vice versa.

ME = Zcritical * Standard error.

We see that the critical value is in the numerator, therefore as the critical value increases, the width will increase.

As the confidence level moves from 80% to 99%, the critical value increases, and therefore the width of the confidence interval also increases.

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