Answer:
Step-by-step explanation:
confidence interval formula is epressed as;
CI = xbar ± z(s/√n)
xbar is the mean = $115
z is the z score at 90% CI = 1.645
s is the standard deviation = 18.10
n is the sample size = 35
Substitute;
CI = 115 ± 1.645(18.10/√35)
CI = 115 ± 1.645(3.0594)
CI = 115± 5.0328
CI = (115-5.0328, 115+5.0328)
CI = (109.97, 120.03)
Hence a 90% confidence interval for the mean electricity bill of all UCF students is 109.97<x<120.03
Using the t-distribution, it is found that the 90% confidence interval for the mean electricity bill of all UCF students is (108.79, 121.21).
We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The information given is:
Sample mean of [tex]\overline{x} = 115[/tex].
Sample standard deviation of [tex]s = 18.1[/tex].
Sample size of [tex]n = 35[/tex].
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 35 - 1 = 34 df, is t = 2.0301.
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 115 - 2.0301\frac{18.1}{\sqrt{35}} = 108.79[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 115 + 2.0301\frac{18.1}{\sqrt{35}} = 121.21[/tex]
The 90% confidence interval for the mean electricity bill of all UCF students is (108.79, 121.21).
A similar problem is given at https://brainly.com/question/15180581
