In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of weekly food expenses is normally shaped.

Step-by-step explanation:

Given that in a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50.

Sample size = 28<30

For a normal distribution to be used, the condition is sample size should be atleast 30 and the sample is strictly drawn at random.

Here we have sample size <30 hence t test can be used as the original distribution is normal.

To construct confidence interval t - distribution should be used