Question

A politician claims that she will receive 60% of the votes in an upcoming election. The results of a simple random sample of 100 voters showed that 52 of those samples will vote for her. (a) Construct a 92% confidence interval to estimate the actual percent of votes she will receive in the upcoming election. (b) If the confidence level was increased, is it likely that her claim would still be supported? Justify without constructing the new confidence interval.

Answer 1

(a) Construct a 92% confidence interval to estimate the actual percent of votes she will receive in the upcoming election.

Confidence interval for Population Proportion is given as below:

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

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 52

n = 100

P = x/n = 52/100 = 0.52

Confidence level = 92%

Critical Z value = 1.7507

(by using z-table)

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

Confidence Interval = 0.52 ± 1.7507* sqrt(0.52*(1 – 0.52)/ 100)

Confidence Interval = 0.52 ± 1.7507* 0.0500

Confidence Interval = 0.52 ± 0.0875

Lower limit = 0.52 - 0.0875 = 0.4325

Upper limit = 0.52 + 0.0875 = 0.6075

Confidence interval = (0.4325, 0.6075)

43.25% < p < 60.75%

(b) If the confidence level was increased, is it likely that her claim would still be supported?

If the confidence level was increased, it is likely that her claim would still be supported, because as we increase the confidence level, the width of the confidence interval will be increase and this increased width or margin of error will be more likely to support her claim.