When is unknown and the sample is of size n 30, there are two methods for comput
ID: 3151437 • Letter: W
Question
When is unknown and the sample is of size n 30, there are two methods for computing confidence intervals for . Method 1:
Use the Student's t distribution with d.f. = n 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method.
Method 2: When n 30, use the sample standard deviation s as an estimate for , and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for . Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution.
Consider a random sample of size n = 31, with sample mean x = 45.0 and sample standard deviation s = 5.6. (a) Compute 90%, 95%, and 99% confidence intervals for using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit (b) Compute 90%, 95%, and 99% confidence intervals for using Method 2 with the standard normal distribution. Use s as an estimate for . Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit
(c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution? Yes. The respective intervals based on the t distribution are longer. No. The respective intervals based on the t distribution are shorter. No. The respective intervals based on the t distribution are longer. Yes. The respective intervals based on the t distribution are shorter. (
d) Now consider a sample size of 71. Compute 90%, 95%, and 99% confidence intervals for using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit
(e) Compute 90%, 95%, and 99% confidence intervals for using Method 2 with the standard normal distribution. Use s as an estimate for . Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit
(f) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
No. The respective intervals based on the t distribution are longer.
No. The respective intervals based on the t distribution are shorter. Yes. The respective intervals based on the t distribution are shorter.
Yes. The respective intervals based on the t distribution are longer.
With increased sample size, do the two methods give respective confidence intervals that are more similar?
As the sample size increases, the difference between the two methods becomes greater.
As the sample size increases, the difference between the two methods remains constant.
As the sample size increases, the difference between the two methods is less pronounced.
Explanation / Answer
Consider a random sample of size n = 31, with sample mean x = 45.0 and sample standard deviation s = 5.6. (a) Compute 90%, 95%, and 99% confidence intervals for using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit
i)
FOR 90% CONFIDENCE:
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 45
t(alpha/2) = critical t for the confidence interval = 1.697260887
s = sample standard deviation = 5.6
n = sample size = 31
df = n - 1 = 30
Thus,
Margin of Error E = 1.707087503
Lower bound = 43.2929125
Upper bound = 46.7070875
Thus, the confidence interval is
( 43.2929125 , 46.7070875 ) [ANSWER]
**************************
ii)
FOR 95% CONFIDENCE:
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 45
t(alpha/2) = critical t for the confidence interval = 2.042272456
s = sample standard deviation = 5.6
n = sample size = 31
df = n - 1 = 30
Thus,
Margin of Error E = 2.054096583
Lower bound = 42.94590342
Upper bound = 47.05409658
Thus, the confidence interval is
( 42.94590342 , 47.05409658 ) [ANSWER]
**************************
iii)
FOR 99% CONFIDENCE:
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 45
t(alpha/2) = critical t for the confidence interval = 2.749995654
s = sample standard deviation = 5.6
n = sample size = 31
df = n - 1 = 30
Thus,
Margin of Error E = 2.76591728
Lower bound = 42.23408272
Upper bound = 47.76591728
Thus, the confidence interval is
( 42.23408272 , 47.76591728 ) [ANSWER]
******************************************************************************************************************************
(b) Compute 90%, 95%, and 99% confidence intervals for using Method 2 with the standard normal distribution. Use s as an estimate for . Round endpoints to two digits after the decimal. 90% 95% 99% lower limit upper limit
i)
FOR 90% CONFIDENCE:
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 45
z(alpha/2) = critical z for the confidence interval = 1.644853627
s = sample standard deviation = 5.6
n = sample size = 31
Thus,
Margin of Error E = 1.654376822
Lower bound = 43.34562318
Upper bound = 46.65437682
Thus, the confidence interval is
( 43.34562318 , 46.65437682 ) [ANSWER]
********************
FOR 95% CONFIDENCE:
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 45
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 5.6
n = sample size = 31
Thus,
Margin of Error E = 1.971311571
Lower bound = 43.02868843
Upper bound = 46.97131157
Thus, the confidence interval is
( 43.02868843 , 46.97131157 ) [ANSWER]
********************
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 45
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 5.6
n = sample size = 31
Thus,
Margin of Error E = 2.59074256
Lower bound = 42.40925744
Upper bound = 47.59074256
Thus, the confidence interval is
( 42.40925744 , 47.59074256 ) [ANSWER]
*******************************************
Hi! Please submit the next part (c onwards) as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.