Question

In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determine whether a normal distribution (Z values) 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 expense is normally shaped.

Answer 1

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

Since in this question we are provided with;

Sample average weekly food expense, $\bar X$ = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be t-distribution because in the question we don't know anything about population standard deviation $(\sigma)$ .

Normal distribution is used when we know population standard deviation $(\sigma)$.

So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;

                P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

Therefore, t-distribution should be used to construct a confidence interval.