Question

The situation is as follows: Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at college. A group of researchers at the Off-Campus Housing department want to estimate the mean monthly rent that unmarried students paid during Winter 2019. During March 2019, they randomly sampled 366 students and found that on average, students paid $348 for rent with a standard deviation of $76. The plot of the sample data showed no extreme skewness or outliers. Calculate a 98% confidence interval estimate for the mean monthly rent of all unmarried students in Winter 2019. QUESTION: What is a 98% confidence interval estimate for the mean monthly rent of all unmarried BYU students in Winter 2019?

1A. State the name of the appropriate estimation procedure.

1B Describe the parameter of interest in the context of the problem.

1C. Name the conditions for the procedure.

1D. Explain how the above conditions are met. (

1E. Write down the confidence level and the t* critical value.

1F. Calculate the margin of error for the interval to two decimal places. Show your work.

1G. Calculate the confidence interval to two decimal places and state it in interval form.

1H. CONCLUDE Interpret your confidence interval in context. Do this by including these three parts in your conclusion:  Level of confidence, Parameter of interest in context, the interval estimate

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =348
standard deviation, s =76
sample size, n =366
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 76/ sqrt ( 366) )
= 3.97
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 365 d.f is 2.337
margin of error = 2.337 * 3.97
= 9.28
III.
CI = x ± margin of error
confidence interval = [ 348 ± 9.28 ]
= [ 338.72 , 357.28 ]
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DIRECT METHOD
given that,
sample mean, x =348
standard deviation, s =76
sample size, n =366
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 365 d.f is 2.337
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 348 ± t a/2 ( 76/ Sqrt ( 366) ]
= [ 348-(2.337 * 3.97) , 348+(2.337 * 3.97) ]
= [ 338.72 , 357.28 ]
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interpretations:
1) we are 98% sure that the interval [ 338.72 , 357.28 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
Answer:
A.
the name of the appropriate estimation procedure is confidence interval for t test single mean
B.
the parameter of interest in the context of the problem
A group of researchers at the Off-Campus Housing department want to estimate the mean monthly rent that
unmarried students paid during Winter 2019
c.
98% confidence interval estimate for the mean monthly rent of all unmarried students in Winter 2019
d.
the above conditions are met for given data mean and stanadard deviation , sample size and confidence
interval
e.
confidence level is 98%
t critical value = 2.337
f.
margin of error =9.28
g.
98% sure that the interval [ 338.72 , 357.28 ]
h.
interpretations:
1) we are 98% sure that the interval [ 338.72 , 357.28 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean