JT = 5x+ 8 CT = 70, and CJ = 7x + 2 Find JT
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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.
<em><u>Since in this question we are provided with;</u></em>
Sample average weekly food expense, = $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 <u>t-distribution</u> because in the question we don't know anything about population standard deviation .
Normal distribution is used when we know population standard deviation .
So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;
P.Q. = ~
Therefore, t-distribution should be used to construct a confidence interval.