So I have two problems below revolving around confidence intervals, and I believe I completed a) successfully. I'm having trouble understanding how to check for b) and would appreciate a push in the right direction.
The public opinion is split 65% against and 35% for increasing taxes to help balance the federal budget. Suppose also that 600 people from the population are selected randomly and interviewed.
a. Suppose that in the sample of 600, 120 people are in favour of increasing tax. Based on this information construct a 95% confidence interval for the true value of the proportion of people who are in favour of increasing taxes and remember you know its actual value is p = 35%.
[p = actual value of the population proportion = 0.35]
p̂ = proportion of the sample in favour = 120/600 = 0.2
(1-p̂) = 0.8
n = sample size = 600
σ(sample) = √[p̂(1-p̂)/600)] = √[0.2*0.8/600] = 0.0163
From the z-table 95% level gives z = 1.96
Margin of error = 1.96*0.0163 = 0.032
0.2-0.032 = 0.168
0.2+0.032 = 0.232
(0.168, 0.232)
b. Was your 95% confidence interval successful in capturing the value p = 35%, which is the true value of the proportion of people who are in favour of increasing taxes? How would you explain this?
Is 0.35 in your calculated confidence interval? No. So your experiment has produced one of the confidence intervals that occur 5% of the time and do not capture the true mean.