A random poll of 800 working men found that 11% had taken on a second job to help pay the bills. a) Estimate the true percentage of men that are taking on second jobs by constructing a 95% confidence interval. b) A pundit on a TV news show claimed that only 8% of working men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data.

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.

Step-by-step 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!!!

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