In: Statistics and Probability
1.
A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 95% confidence interval for the population mean.
Select one:
A. 24.904 to 31.096
B. 25.905 to 32.096
C. 26.324 to 29.432
D. None of the above answers is correct
2.
A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 99% confidence interval for the population mean.
Select one:
a. 23.80-32.20
b. 25.43-30.57
c. 24.9-31.1
3.
A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 90% confidence interval for the population mean.
Select one:
a. 24.90-31.10
b. 25.43-30.57
c. 22.0-32.0
4.
A sample of 491 lunch customers was taken at a restaurant. The average amount of time the customers stayed at the restaurant averaged 33 minutes. From past experience, the standard deviation is known to be 10 minutes. What is the value for the point estimate for a confidence interval for the mean?
Select one:
a. 33
b. 10
c. 4.91
5.
A sample of 491 lunch customers was taken at a restaurant. The average amount of time the customers stayed at the restaurant averaged 33. From past experience, the standard deviation is known to be 10. Construct a 99% confidence interval
Select one:
a. 28.314-37.686
b. 30.314-36.686
c. 31.838-34.162
6.
A sample of 491 lunch customers was taken at a restaurant. The average amount of time the customers stayed at the restaurant averaged 33. From past experience, the standard deviation is known to be 10. Construct a 95% confidence interval.
Select one:
a. 31.838-34.162
b. 32.115-33.885
c. 32.258-33.742
7.
A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 90% confidence interval for the population mean.
Select one:
a. 24.90-31.10
b. 25.43-30.57
c. 22.0-32.0
Question 1
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 25- 1 ) = 2.064
28 ± t(0.05/2, 25 -1) * 7.5/√(25)
Lower Limit = 28 - t(0.05/2, 25 -1) 7.5/√(25)
Lower Limit = 24.904
Upper Limit = 28 + t(0.05/2, 25 -1) 7.5/√(25)
Upper Limit = 31.096
95% Confidence interval is ( 24.904 , 31.096 )
A. 24.904 to 31.096
Question 2
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 25- 1 ) = 2.797
28 ± t(0.01/2, 25 -1) * 7.5/√(25)
Lower Limit = 28 - t(0.01/2, 25 -1) 7.5/√(25)
Lower Limit = 23.80
Upper Limit = 28 + t(0.01/2, 25 -1) 7.5/√(25)
Upper Limit = 32.20
99% Confidence interval is ( 23.80 , 32.20 )
a. 23.80-32.20
Question 3
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 25- 1 ) = 1.711
28 ± t(0.1/2, 25 -1) * 7.5/√(25)
Lower Limit = 28 - t(0.1/2, 25 -1) 7.5/√(25)
Lower Limit = 25.43
Upper Limit = 28 + t(0.1/2, 25 -1) 7.5/√(25)
Upper Limit = 30.57
90% Confidence interval is ( 25.43 , 30.57 )
b. 25.43-30.57
Question 4
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Point Estimate = X̅ = 33
Question 5
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
33 ± Z (0.01/2 ) * 10/√(491)
Lower Limit = 33 - Z(0.01/2) 10/√(491)
Lower Limit = 31.838
Upper Limit = 33 + Z(0.01/2) 10/√(491)
Upper Limit = 34.162
99% Confidence interval is ( 31.838 , 34.162 )
c. 31.838-34.162
Question 6
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
33 ± Z (0.05/2 ) * 10/√(491)
Lower Limit = 33 - Z(0.05/2) 10/√(491)
Lower Limit = 32.115
Upper Limit = 33 + Z(0.05/2) 10/√(491)
Upper Limit = 33.885
95% Confidence interval is ( 32.115 , 33.885 )
b. 32.115-33.885
Question 7
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 25- 1 ) = 1.711
28 ± t(0.1/2, 25 -1) * 7.5/√(25)
Lower Limit = 28 - t(0.1/2, 25 -1) 7.5/√(25)
Lower Limit = 25.43
Upper Limit = 28 + t(0.1/2, 25 -1) 7.5/√(25)
Upper Limit = 30.57
90% Confidence interval is ( 25.43 , 30.57 )
b. 25.43-30.57