Question

A random sample of 58 randomly selected people were asked how much they spend on their grocery bill each week. The mean was $115.36 with a standard deviation of $16.6. Find a 95% confidence interval for the true population mean amount spent on groceries each week. Report the answer accurate to four decimal places.

Answer 1

Answer)

As the population s.d is not mentioned here and we are given with the sample s.d as the best estimate

We will use t distribution to estimate the interval

Degrees of freedom is = n-1 = 57

For 57 dof and 95% confidence level critical value t from t table is = 2.2

Margin of error (MOE) = t*s.d/√n = 2.2*16.6/√58 = 4.3647

Interval is given by

(Mean - MOE, Mean + MOE)

CI [110.9953, 119.7247].

You can be 95% confident that the population mean (μ) falls between 110.9953 and 119.7247.