Answer:
90% confidence interval for the population average amount of purchases is [$4.54 , $6.71].
Explanation:
We are given the randomly sampled 24 purchases from several convenience stores in suburban Long Island and tabulated the amounts to the nearest dollar;
2, 11, 8, 7, 9, 3, 5, 4, 2, 1, 10, 8, 14, 7, 6, 3, 7, 2, 4, 1, 3, 6, 8, 4
Assume that the population standard deviation is 3.23 and the population is normally distributed.
Firstly, the Pivotal quantity for 90% confidence interval for the population average is given by;
P.Q. =
~ N(0,1)
where,
= sample average amount of purchases =
= $5.625
= population standard deviation = $3.23
n = sample of purchases = 24
= population average amount of purchases
Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.
So, 90% confidence interval for the population average,
is ;
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 <
< 1.645) = 0.90
P(
<
<
) = 0.90
P(
<
<
) = 0.90
90% confidence interval for
= [
,
]
= [
,
]
= [4.54 , 6.71]
Therefore, 90% confidence interval for the population average amount of purchases is [$4.54 , $6.71].