Confidence interval using T scores

Question

A recent political poll by the Salt Lake Tribune shows Frmr. Gov. Mitt Romney (R) leading his Utah Senate race with Jenny Wilson (D) 60% to 14% with 26% undecided. The poll sampled 100 likely Utah voters. Using T-scores, construct a 95% confidence interval capturing the true population proportion of Wilson supporters.

Answer 1

To construct a 95% confidence interval for the proportion of Wilson supporters in the population, we need to use the following formula:

CI = p̂ ± tα/2 * sqrt(p̂(1-p̂)/n)

where:

p̂ is the sample proportion of Wilson supporters (14% or 0.14)

tα/2 is the critical value of the t-distribution with n-1 degrees of freedom and α = 0.05/2 = 0.025 (since we want a 95% confidence interval, which leaves 2.5% in each tail)

n is the sample size (100)

First, we need to find the value of tα/2. We can look this up in a t-distribution table or use a calculator. For a two-tailed test with 99 degrees of freedom and α = 0.025, the value of tα/2 is approximately 1.984.

Now we can plug in the values and calculate the confidence interval:

CI = 0.14 ± 1.984 * sqrt(0.14*(1-0.14)/100) = 0.14 ± 0.168

Therefore, the 95% confidence interval for the proportion of Wilson supporters in the population is (0.14 - 0.168, 0.14 + 0.168) or approximately (0, 0.308). This means we can be 95% confident that the true proportion of Wilson supporters in the population falls between 0% and 30.8%.