In a sample of 200 adults, standard deviation of $14.50, 95% confidence interval and $35 weekly expenditure

Question

For 200 adults surveyed standard deviation of $14.50 and $35 average weekly expenditure, what is the 95% confidence interval?

Answer 1

Mean, μ=$35, SD, σ=$14.50, n=200. Z=1.96 for 2-tailed 95% CI.

|(X-μ)/σ|=1.96, X=μ±1.96σ=35±28.42, giving a CI of ($6.58,$63.42).

This means that in 95% of cases the average weekly expenditure will be between $6.58 and $63.42.