Question

19 multiple choice 4 points select the most appropriate response. 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 approximately normal. cannot use normal distribution or t - distribution. use a t - distribution. use the normal distribution.

Explanation:

Step1: Recall the conditions for using normal and t - distribution

For constructing a confidence interval for the population mean \(\mu\):

• If the population standard deviation \(\sigma\) is known and the population is normal or \(n\geq30\) (by Central Limit Theorem), we use the normal distribution \(z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\).
• If the population standard deviation \(\sigma\) is unknown, and the population is normal (or \(n\) is large, \(n\geq30\)), we use the t - distribution \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(s\) is the sample standard deviation.

Step2: Analyze the given information

We are given that \(n = 28\) (sample size), \(\bar{x}=95.60\) (sample mean), \(s = 22.50\) (sample standard deviation). The population standard deviation \(\sigma\) is unknown. But we are told that the distribution of weekly food expenses is approximately normal.

Since \(\sigma\) is unknown and the population is approximately normal (and \(n<30\)), we use the t - distribution.

Answer:

Use a t - distribution.