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

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A simple random sample of 60 items resulted in a sample mean of
80. The population standard deviation is
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(Round your answers to two decimal places.)
to
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Assume that the same sample mean was obtained from a sample of
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