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

A simple random sample of 70 items resulted in a sample mean of
60. The population standard deviation is σ = 15.
(a) Compute the 95% confidence interval for the population mean.
(Round your answers to two decimal places.)
_________ to _________
(b) Assume that the same sample mean was obtained from a sample
of 140 items. Provide a 95% confidence interval for the population
mean. (Round your answers to two decimal places.)
_________ to _________
(c) What is the...

A simple random sample of 30 items resulted in a sample mean of
25. The population standard deviation is 7 . Round your answers to
two decimal places
a. What is the standard error of the mean,
?
b. At 95% confidence, what is the margin of
error?

A simple random sample of 30 items resulted in a sample mean of
30. The population standard deviation is 15.
a. Compute the 95% confidence interval for the
population mean. Round your answers to one decimal place.
( , )
b. Assume that the same sample mean was
obtained from a sample of 120 items. Provide a 95% confidence
interval for the population mean. Round your answers to two decimal
places.
( , )

A simple random sample of 60 items resulted in a sample mean of
80. The population standard deviation is
σ = 5.
(a)
Compute the 95% confidence interval for the population mean.
(Round your answers to two decimal places.)
to
(b)
Assume that the same sample mean was obtained from a sample of
120 items. Provide a 95% confidence interval for the population
mean. (Round your answers to two decimal places.)
to

A simple random sample of 60 items resulted in a sample mean of
80. The population standard deviation is
σ = 5.
(a)
Compute the 95% confidence interval for the population mean.
(Round your answers to two decimal places.)
to
(b)
Assume that the same sample mean was obtained from a sample of
120 items. Provide a 95% confidence interval for the population
mean. (Round your answers to two decimal places.)
to

A simple random sample of 60 items resulted in a sample mean of
65. The population standard deviation is 16. a. Compute the 95%
confidence interval for the population mean (to 1 decimal). ( , )
b. Assume that the same sample mean was obtained from a sample of
120 items. Provide a 95% confidence interval for the population
mean (to 2 decimals). ( , ) c. What is the effect of a larger
sample size on the margin of...

A simple random sample of 60 items resulted in a sample mean of
65. The population standard deviation is 16. a. Compute the 95%
confidence interval for the population mean (to 1 decimal). ( , )
b. Assume that the same sample mean was obtained from a sample of
120 items. Provide a 95% confidence interval for the population
mean (to 2 decimals). ( , ) c. What is the effect of a larger
sample size on the margin of...

A simple random sample of 60 items resulted in a sample mean of
95. The population standard deviation is 13.
a. Compute the 95% confidence interval for the population mean
(to 1 decimal).
( , )
b. Assume that the same sample mean was obtained from a sample
of 120 items. Provide a 95% confidence interval for the population
mean (to 2 decimals).
( , )
c. What is the effect of a larger sample size on the margin of
error?
SelectIt increasesIt decreasesIt...

A simple random sample of 60 items resulted in a sample mean of
10. The population standard deviation is 20.
Compute the 95% confidence interval for the population mean.
Round to 1 decimal place.
Assume that the same sample mean was obtained from a sample of
120 items. Provide a 95% confidence interval for the population
mean. Round to 2 decimal places.
What is the effect of a larger sample size on the interval
estimate? Larger sample provides a larger...

A simple random sample of 60 items resulted in a sample mean of
63. The population standard deviation is 12.
a. Compute the 95% confidence interval for the population mean
(to 1 decimal).
( , )
b. Assume that the same sample mean was obtained from a sample
of 120 items. Provide a 95% confidence interval for the population
mean (to 2 decimals).
( , )

ADVERTISEMENT

ADVERTISEMENT

Latest Questions

- A die has been "loaded" so that the probability of rolling any even number is 5...
- A population of values has a normal distribution with μ = 96.2 μ = 96.2 and...
- You work for SneauxCeauxne, the Cajun-inspired crushed-ice dessert company. You’ve collected data on monthly sales (S,...
- Taxation Questions Q1: Anna converted cryptocurrency into $27,200 Australian dollars in October 2019. To complete the...
- Discussion: The Strategy of Cultivating Interdependence All of life is an opportunity to think, relate, and...
- People in the aerospace industry believe the cost of a space project is a function of...
- O’Brien Company manufactures and sells one product. The following information pertains to each of the company’s...

ADVERTISEMENT