Question

In a recent issue of Time magazine, 15 of 68 pages contained advertizing. Count every page that includes any advertising. Find the proportion of pages that include advertising.

Based on this sample result, construct a 95% confidence interval estimate of the percentage of all popular magazine pages that have advertising. (See section 7-2 of the textbook for examples of such confidence intervals).

Give a description of what this confidence interval tells you about magazine advertising.

What would change about your confidence interval if you changed the confidence level to 99%?

Answer 1

p= 0.22...................(proportion)

q= 0.78

n= 68

alpha=0.05 then Z(alpha/2)= 1.96

Margin of error E=Z(alpha/2)*sqrt(p*q/n)

=1.96*sqrt(0.22*0.78/68)

=0.0984600

95% Confidence interval for population proportion =(p-E,p+E)

rounded lower bound= 0.1215

rounded upper bound= 0.3185

there is a 95% chance that population proportion fall between 12.15% and 31.85%

p= 0.22

q= 0.78

n= 68

alpha=0.01 then Z(alpha/2)= 2.576

Margin of error E=Z(alpha/2)*sqrt(p*q/n)

=2.576*sqrt(0.22*0.78/68)

=0.1294

99% Confidence interval for population proportion =(p-E,p+E)

rounded lower bound= 0.0906

rounded upper bound= 0.3494

There is a 99% chance that population proportion fall between 9.06% and 34.94%

As the confidence level increased from 95% to 99% , it become wider on normal curve.

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