A simple random sample of 31 observations is derived from a normally distributed

| August 30, 2017

Q1
A simple random sample of 31 observations is derived from a normally distributed population with a known standard deviation of 2.3. Use Table 1.

a.
Is the condition that is normally distributed satisfied?

Yes
No

b.
Compute the margin of error with 94% confidence. (Do not round intermediate calculations. Round “z” value to 3 decimal places and final answer to 2 decimal places.)

Margin of error

c.
Compute the margin of error with 97% confidence. (Do not round intermediate calculations. Round “z” value to 3 decimal places and final answer to 2 decimal places.)

Margin of error

d.
Which of the two margins of error will lead to a wider interval?

The margin of error with 2.3% confidence.
The margin of error with 97% confidence.

Q2.
Consider a normal population with an unknown population standard deviation. A random sample results in = 52.23 and s2 = 22.09. Use Table 2.

a.
Compute a 88% confidence interval for ? if and s2 were obtained from a sample of 15 observations. (Do not round intermediate calculations. Round “t?/2,df” value to 3 decimal places and final answers to 2 decimal places.)

Confidence interval
to

b.
Compute a 88% confidence interval for ? if and s2 were obtained from a sample of 24 observations. (Do not round intermediate calculations. Round “t?/2,df” value to 3 decimal places and final answers to 2 decimal places.)

Confidence interval
to

Q3.
A recent study found that consumers are making average monthly debt payments of $983 (Experian.com, November 11, 2010). The accompanying table shows the average debt payments for 26 metropolitan areas. Use Table 2.

City
Debt Payments
Washington, D.C.
$1,285
Seattle
1,135
Baltimore
1,133
Boston
1,133
Denver
1,104
San Francisco
1,098
San Diego
1,076
Sacramento
1,045
Los Angeles
1,024
Chicago
1,017
Philadelphia
1,011
Minneapolis
1,011
New York
989
Atlanta
986
Dallas
970
Phoenix
957
Portland
948
Cincinnati
920
Houston
889
Columbus
888
St. Louis
886
Miami
867
Detroit
832
Cleveland
812
Tampa
791
Pittsburgh
763
SOURCE: .Experian.com”>http://www.Experian.com, November 11, 2010.

Click here for the Excel Data File

a.
Use Excel to calculate the mean and standard deviation for debt payments. (Do not round intermediate calculations. Round your answers to 2 decimal places.)

Sample mean

Sample standard deviation

b.
Construct a 90% and a 95% confidence interval for the population mean. (Do not round intermediate calculations. Round “t?/2,df” value to 3 decimal places and final answers to 2 decimal places.)

Confidence Level
Confidence Interval
90%

to

95%

to

Q4.
One in five 18-year-old Americans has not graduated from high school (The Wall Street Journal, April 19, 2007). A mayor of a northeastern city comments that its residents do not have the same graduation rate as the rest for the country. An analyst from the Department of Education decides to test the mayor’s claim. In particular, she draws a random sample of 80 18-year-olds in the city and finds that 20 of them have not graduated from high school. Use Table 1.

a.
Compute the point estimate for the proportion of 18-year-olds who have not graduated from high school in this city. (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Point estimate

b.
Use this point estimate to derive a 95% confidence interval for the population proportion. (Do not round intermediate calculations. Round “z” value to 2 decimal places and final answers to 3 decimal places.)

Confidence interval
to

Q5.
An analyst from an energy research institute in California wishes to precisely estimate a 95% confidence interval for the average price of unleaded gasoline in the state. In particular, she does not want the sample mean to deviate from the population mean by more than $0.06. What is the minimum number of gas stations that she should include in her sample if she uses the standard deviation estimate of $0.26, as reported in the popular press? Use Table 1. (Do not round intermediate calculations. Round “z” value to 2 decimal places. Round up your answer to the nearest whole number.)

Minimum number of gas stations

Q6.
A simple random sample of 37 observations is derived from a normally distributed population with a known standard deviation of 6.4. Use Table 1.

a.
Is the condition that is normally distributed satisfied?

Yes
No

b.
Compute the margin of error with 99% confidence. (Round intermediate calculations to 4 decimal places, “z” value and final answer to 2 decimal places.)

Margin of error

c.
Compute the margin of error with 95% confidence. (Round intermediate calculations to 4 decimal places, “z” value and final answer to 2 decimal places.)

Margin of error

Q7.
Consider a population with a known standard deviation of 11.2. In order to compute an interval estimate for the population mean, a sample of 43 observations is drawn. Use Table 1.

a.
Is the condition that is normally distributed satisfied?

Yes
No

b.
Compute the margin of error at a 90% confidence level. (Round your intermediate calculations to 4 decimal places. Round “z” value and final answer to 2 decimal places.)

Margin of error

c.
Compute the margin of error at a 90% confidence level based on a larger sample of 470 observations. (Round your intermediate calculations to 4 decimal places. Round “z” value and final answer to 2 decimal places.)

Margin of error

d.
Which of the two margins of error will lead to a wider confidence interval?

90% confidence with n = 43.
90% confidence with n = 470.

Q8.
A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. They ask their realtor friend for help and she informs them that the last 16 houses that sold in their neighborhood took an average time of 250 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 55 days. Use Table 1.

a.
What assumption regarding the population is necessary for making an interval estimate for the population mean?

Assume that the central limit theorem applies.
Assume that the population has a normal distribution.

b.
Construct a 90% confidence interval for the mean sale time for all homes in the neighborhood. (Round intermediate calculations to 4 decimal places, “z” value and final answer to 2 decimal places.)

Confidence interval
to

Q9.
Find t?,df from the following information. Use Table 2. (Round your answers to 3 decimal places.)

t?,df
a. ? = 0.005 and df = 25

b. ? = 0.20 and df = 25

c. ? = 0.005 and df = 10

d. ? = 0.20 and df = 10

Q10.
A random sample of 26 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 125.4 and 26.50, respectively. Assume that the population is normally distributed. Use Table 2.

a.
Construct a 95% confidence interval for the population mean. (Round intermediate calculations to 4 decimal places, “t” value to 3 decimal places, and final answers to 2 decimal places.)

Confidence interval

to

b.
Construct a 99% confidence interval for the population mean. (Round intermediate calculations to 4 decimal places, “t” value to 3 decimal places, and final answers to 2 decimal places.)

Confidence interval
to

Q11.
Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 15, 25, 12, 18, 28, 17, 14, 24. Use Table 2.

a.
Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to 4 decimal places, “sample mean” to 3 decimal places and “sample standard deviation” to 2 decimal places.)

Sample mean

Sample standard deviation

b.
Construct the 95% confidence interval for the population mean. (Round “t” value to 3 decimal places, and final answers to 2 decimal places.)

Confidence interval
to

c.
Construct the 99% confidence interval for the population mean. (Round “t” value to 3 decimal places, and final answers to 2 decimal places.)

Confidence interval
to

Q12.
The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served was 200, 145, 125, 260, 220, and 100. She believes that the number of customers served on weekdays follows a normal distribution. Construct a 95% confidence interval for the average number of customers served on weekdays. Use Table 2. (Round intermediate calculations to 4 decimal places, “sample mean” and “sample standard deviation” to 2 decimal places and “t” value to 3 decimal places, and final answers to 2 decimal places.)

Confidence interval
to

Q13.
A random sample of 170 observations results in 119 successes. Use Table 1.

a.
Construct a 99% confidence interval for the population proportion of successes. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answers to 3 decimal places.)

Confidence interval
to

b.
Construct a 99% confidence interval for the population proportion of failures. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answers to 3 decimal places.)

Confidence interval
to

Q14.
An economist reports that 882 out of a sample of 1,800 middle-income American households actively participate in the stock market.Use Table 1.

a.
Construct a 90% confidence interval for the proportion of middle-income Americans who actively participate in the stock market. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answers to 3 decimal places.)

Confidence interval
to

Q15.
An accounting professor is notorious for being stingy in giving out good letter grades. In a large section of 130 students in the fall semester, she gave out only 5% As, 28% Bs, 38% Cs, and 29% Ds and Fs. Assuming that this was a representative class, compute a 95% confidence interval of the probability of getting at least a B from this professor. Use Table 1. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answers to 3 decimal places.)

Confidence interval
to

Q16.
In the planning stage, a sample proportion is estimated as = 18/30 = 0.60. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error E = 0.11. What happens to n if you decide to estimate p with 90% confidence? Use Table 1. (Round intermediate calculations to 4 decimal places and “z” value to 2 decimal places. Round up your answers to the nearest whole number.)

Confidence Level
n
95%

90%

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