You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a sample mean of 573 and a sample standard deviation of 22. Calculate the upper confidence limit of a 90% confidence interval for the population mean of rent paid. If necessary, round your answer to four decimal places. HINT: For a confidence interval of [a, b] we refer to "a" as the "lower confidence limit" and we refer to "b" as the "upper confidence limit". g

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

represent the sample mean

population mean (variable of interest)

s=22 represent the sample standard deviation

n=36 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

  (1)

The degrees of freedom are given by:

The Confidence level is 0.90 or 90%, the significance would be and , and the critical value for this case would be

And replacing we got:

And the 90% confidence interval would be between [566.8033 and 579.1967]

step-by-step explanation:

i need to answer questions so i dont know but i think its the first one because it goes up but a consent rate

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*