A random sample of 35 UCF students has a mean electricity bill of $115. Assume the population standard deviation is $18.10. Construct a 90% confidence interval for the mean electricity bill of all UCF students. Round final answer to two decimal places.

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 .

Sample standard deviation of .

Sample size of .

The confidence interval is:

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:

The 90% confidence interval for the mean electricity bill of all UCF students is (108.79, 121.21).

A similar problem is given at link

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

the correct answer is you spend $55.78 more in a brand store that buy ing in a normal store.

step-by-step explanation:

brand = spend   $ 135.85 total

store = spend $80.07 total

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!