Compute a 95% confidence interval for the mean family expenditure?

Expenditure on one category of food is estimated on the basis of a family expenditure survey. The amounts spent are surpassed to vary around an unknown mean, with a standard deviation of 6 dollars. For a sample of 80 families, the average expenditure is 43 dollars.





a) Compute a 95% confidence interval for the mean family expenditure?





b) How large of a sample is required to estimated the mean expenditure to within 50 cents?





c) Do we need to assume that the individual expenditure follow a normal distribution?

Compute a 95% confidence interval for the mean family expenditure?
In this case n = 80, x-bar = 43, s = 6





Standard error of the mean = 6 /√80 = 0.67





The 95% confidence interval = (μ - 1.96 x SE) - (μ + 1.96 x SE), and 1.96 x 0.67 = 1.32





Hence (43 - 1.32) %26lt; μ %26lt; (43 + 1.32), i.e. 41.68 %26lt; μ %26lt; 44.32





b. (This corrects an error in my original answer to this question.)





To get an estimate to within 50 cents with a 95% confidence interval it is necessary to apply the z-value as determined above to the standard error of the mean.





In other words, we require that 1.96 x 6 / √n = 0.5





So √n = (1.96 x 6) / 0.5 = 23.52 or n = 553





So with the same sample mean, for example, this would give the confidence interval as





42.50 %26lt; μ %26lt; 43.50


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c. No. The sample means are normally distributed in all cases, irrespective of the shape of the population distribution.