Question

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. o a. use normal distribution. o b. use the t - distribution. o c. cannot use normal distribution or t - distribution.

Explanation:

Step1: Check sample size and standard deviation type

The sample size \(n = 28\) (small - sample, \(n<30\)). The population standard deviation \(\sigma\) is unknown (we are given the sample standard deviation \(s = 22.50\)).

Step2: Recall distribution selection rules

When the population is normally distributed, for small - samples (\(n < 30\)) and unknown population standard deviation, the \(t\) - distribution is used. The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2}\) is the critical value from the \(t\) - distribution.

Answer:

B. Use the t - distribution.