Question

The proportion of adult women in a certain geographical region is approximately 51​%. A marketing survey telephones 400 people at random. Complete parts a through e below.

How would the confidence interval change if the confidence level had been 90​% instead of 98​%?

Answer 1

Solution:

First of all, we have to find the 90% confidence interval for the population proportion of adult women in a certain geographical region.

Confidence interval = P ± Z*sqrt(P*(1 – P)/n)

We are given

Sample proportion = P = 0.51

Sample size = n = 400

Confidence level = 90%

Critical Z value = 1.6449 (by uisng z-table)

Confidence interval = P ± Z*sqrt(P*(1 – P)/n)

Confidence interval = 0.51 ± 1.6449*sqrt(0.51*(1 – 0.51)/400)

Confidence interval = 0.51 ± 1.6449* 0.0250

Confidence interval = 0.51 ± 0.0411

Lower limit = 0.51 - 0.0411 = 0.4689

Upper limit = 0.51 + 0.0411 = 0.5511

Confidence interval = (0.4689, 0.5511)

Now, we have to find a 98% confidence interval for a population proportion.

Confidence interval = P ± Z*sqrt(P*(1 – P)/n)

We are given

Sample proportion = P = 0.51

Sample size = n = 400

Confidence level = 98%

Critical Z value = 2.3263 (by using z-table)

Confidence interval = P ± Z*sqrt(P*(1 – P)/n)

Confidence interval = 0.51 ± 2.3263*sqrt(0.51*(1 – 0.51)/400)

Confidence interval = 0.51 ± 2.3263* 0.0250

Confidence interval = 0.51 ± 0.0581

Lower limit = 0.51 – 0.0581 = 0.4519

Upper limit = 0.51 + 0.0581 = 0.5681

Confidence interval = (0.4519, 0.5681)

If we use a 90% confidence interval for population proportion instead of a 98% confidence interval for the population proportion, then the width of the confidence interval becomes smaller. This means, less confidence level results in narrower confidence level if all other variables or values are same.