Answer:
a) 95% Confidence Interval = (8.832%, 13.168%)
b) The 8% claim for the pundit falls outside the range of the confidence interval, hence, it isn't a very plausible claim given the poll data.
Explanation:
Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample proportion) ± (Margin of error)
Sample proportion = 11% = 0.11
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the sample proportion)
Critical value will be obtained using the z-distribution. This is because the sample size is large enough for the t-distributoon valur to approximate the z-distribution value
Critical value for 95% confidence = 1.960 (from the z-tables)
Standard error = σₓ = √[p(1-p)/n]
where
p = sample proportion = estimated to be 11% = 0.11
n = Sample size = 800
σₓ = √[p(1-p)/n]
σₓ = √[0.11×0.89/800]
σₓ = 0.0110623234 = 0.01106
95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]
CI = 0.11 ± (1.960 × 0.01106)
CI = 0.11 ± 0.02168
95% Confidence Interval = (0.08832, 0.13168)
95% Confidence Interval = (8.832%, 13.168%)
b) The 8% claim for the pundit falls outside the range of the confidence interval, hence, it isn't a very plausible claim given the poll data.
Hope this Helps!!!