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Margin of Error Calculations

1. The width of a confidence interval is always twice what value?
2. Suppose = 29:47 and x = 26:54. What is the sampling error?
3. Let c = 0:99, s = 5:6, and n = 87. What is the margin of error, E?
4. Construct the confidence interval for the population mean, .
c = 0:90, x = 58:62, s = 1:84, and n = 65.
Use the confidence interval, (36:482; 41:352), to answer questions 5 and 6.
5. What is the margin of error, E?
6. What is the sample mean, x?
7. Let c = 0:99, s = 3:7, and E = 2:2. What is the minimum sample size, n, needed to estimate? (Assume that s had been found from a preliminary sample that had more than 30 members.)
8. Repair Costs: Refrigerators In a random sample of 87 refrigerators, the mean repair cost
was $195:00 and the standard deviation was $25:50. Construct a 95% confidence interval for
the population mean repair cost.
9. Let c = 0:98 and n = 22. Find the critical value, tc.
10. Let c = 0:80, s = 5, and n = 14. Find the margin of error, E.
Use the information below to answer questions 11-13. Round your results to 3 decimal
places for this problem.
Let c = 0:90, x = 48:51, s = 2:15, and n = 19.
11. Construct a confidence interval for the population mean using the t-distribution.
12. Construct the confidence interval for the population mean that you would get if you incorrectly used a normal distribution.
13. Which interval is wider?
Use the information below to answer questions 14 and 15.
Repair Costs: Computers In a random sample of 12 computers, the mean repair cost was
$149:50 and the standard deviation was $37:89. Assume the repair cost is normally distributed.
14. Use a t-distribution to construct a 98% confidence interval for the population mean .
15. What is the margin of error, E, of ?
Use the table below to answer questions 16-18.
Incomes The monthly incomes for 14 randomly selected people, each with a bachelor's degree. Assume that the population of the monthly incomes is normally distributed.
3885.90 4533.40 6986.70 5100.30 7028.10 7915.40 5076.60
7150.40 5643.50 4628.88 3737.68 6030.58 7968.10 6372.27
16. What is the sample mean, x?
17. What is the sample standard deviations?
18. Construct a 95% confidence interval for the population mean .
Use the information below to answer questions 19 and 20.
Prices of Toasters You took a random sample of 15 two-slice toasters and found the mean
price was $48:89 and the standard deviation was $18:76. Assume the prices are normally
distributed.
19. Which distribution, the t-distribution or the normal distribution, should be used for this problem? State your reason why. If neither distribution can be used, explain why.
20. If you decided to use either the t-distribution or the normal distribution in question 19, construct a 95% confidence interval for the population mean. If you decided that neither distribution could be used, write "not applicable" for your answer.
Use the information below to answer questions 21 to 25.
Let p be the population proportion for the given condition.
Savings In a survey of 5; 208 U.S. adults, 3; 872 say they have saved part of their salary over
the last 12 months.
21. What is the point estimate, p^, for p?
22. What is the point estimate, q^, for q?
23. Construct the 95% confidence interval for the population proportion p.
24. Construct the 99% confidence interval for the population proportion p.
25. Which interval is wider?
Use the information below to answer questions 26-28.
Online Service Usage You are a travel agent and wish to estimate, with 94% confidence, the
proportion of vacationers who use an online service or the internet to make travel reservations.
Your estimate must be accurate to within 4% of the population proportion. (Note: Please do
not use c = 95% to estimate this answer).
26. No preliminary estimate is available. Find the minimum sample size needed.
27. Find the minimum sample size needed, using a prior study that found that 30% of the respondents said they used an online service or the internet to make travel reservations.
28. What do conclude from the results found in questions 26 and 27?
Use the information below to answer questions 29-34.
A survey was conducted to determine if adults in the U.S., France, and Germany believed
humans contribute to global warming. Of 3000 adults in the U.S., 65% believed humans
contribute to global warming, of 1325 adults in France, 79% believed humans contribute to
global warming, and of 1335 adults in Germany, 91% believed humans contribute to global
warming. (Note: Two proportions may possibly be equal if their confidence intervals overlap.)
29. Construct a 99% confidence interval for the proportion of adults from the U.S. who say that
humans contribute to global warming.
30. Construct a 99% confidence interval for the proportion of adults from France who say that
humans contribute to global warming.
31. Construct a 99% confidence interval for the proportion of adults from Germany who say that
humans contribute to global warming.
32. Based on your results for questions 29 and 30, is it possible that the proportion for adults in
the U.S. may equal the proportion of adults in France?
33. Based on your results for questions 30 and 31, is it possible that the proportion for adults in
France may equal the proportion of adults in Germany?
34. Based on your results for questions 29 and 31, is it possible that the proportion for adults in
the U.S. may equal the proportion of adults in Germany

Use the following information to answer questions 1 and 2. Paint Sprayer A company uses an
automated pint sprayer to apply paint to metal furniture. The company sets the sprayer to apply
paint one-mil (1/1000 of an inch) thick. (Note: For the problems below, you must show both sample
size calculations and your conclusion).
1. The company wants to estimate the mean thickness the sprayer is applying paint within 0:0425
mil. Determine the minimum sample size required to construct a 90% confidence interval for the
population mean. Assume the population standard deviation is 0:15 mil.
2. Repeat question 1 using an error tolerance of 0:02125 mil. Which error tolerance requires a larger
samples size? Use the following information to answer questions 3 and 4. Mini-soccer Balls A soccer ball
manufacturer wants to estimate the mean circumference of mini-soccer balls within 0.15 inch. Assume that the population of circumferences is normally distributed. (Note: For the problems below,
you must show both sample size calculations and your conclusion.)
3. Determine the minimum sample size required to construct a 99% confidence interval for the population mean. Assume the population standard deviation is 0:20 inch.
4. Repeat problem 3 using a standard deviation of 0:10 inch. Which standard deviation requires a larger
sample size? Explain.
For questions 5 and 6, you are given a data sample. Construct a 95% confidence interval for
the population mean.
5. Airfare A random sample of airfare prices (in dollars) for a one-way ticket airfare from Chicago, IL
to Minneapolis, MN.
Key: 8j1 = 81
18 1 1 1 1
8 6 6 6 9 9 9 9 9
9 4 4 4 4 4
9 8 8 9 9
10 3 3 3 4 4
10 9 9 9 9
11 5 5 5 5
11 9 9
12 4
6. Annual Precipitation A random sample of the annual precipitation (in inches) for Nome, Alaska.
18.31 19.87 19.76 17.10 19.06 14.93 9.08 20.66 12.29
20.14 24.38 22.15 24.25 13.67 20.80 14.30 7.39 20.09
9.93 14.97 17.13 10.44 22.06 19.25 14.92 13.05 14.17
15.46 16.27 15.23 13.43 17.62 17.49
7. Waste Recycled In a random sample of 12 adults from the United States, the mean waste recycled
per person per day was 1.46 pounds and the standard deviation was 0.28 pound. Construct a 90%
con?dence interval for the population mean .
8. Repeat problem 7, assuming the same statistics came from a sample size of 600. Compare the
results.
9. Light Bulb Manufacturing A company manufactures light bulbs. The company wants the bulbs
to have a mean life span of 1000 hours. This average is maintained by periodically testing random
samples of 16 light bulbs. If the t-value falls between t0:99 and t0:99, the company will be satisfied
that it is manufacturing acceptable light bulbs. A sample of 16 light bulbs is randomly selected
and tested. The mean life span of the sample is 1015 hours and the standard deviation is 25 hours.
Assume the life spans are approximately normally distributed. Is the company making acceptable
light bulbs? Explain your reasoning.
Use the following information to answer questions 10 - 12.Vacation In a survey of 1003 U.S.
adults, 110 say they would go on vacation to Europe if cost did not matter.
10. Construct the 95% confidence interval for the population proportion p.
11. Construct the 99% confidence interval for the population proportion p.
12. From problems 10 and 11, which interval is wider?
13. In a survey of 1001 U.S. adults, 27% said that they smoked a cigarette in the past week. The
survey's margin of error is plus or minus 3%. Translate this information into a confidence interval
for p. Approximate the value of the level of confidence c.

Solution Preview

1. The width of a confidence interval is always twice what value?
The width of confidence interval is always twice of margin of error.
2. Suppose = 29:47 and x = 26:54. What is the sampling error?
Error=26.54-29.47=-2.93
3. Let c = 0:99, s = 5:6, and n = 87. What is the margin of error, E?
The critical value for 99% confidence interval is 2.576.
Margin of error E=2.576*5.6/sqrt(87)=1.55
4. Construct the confidence interval for the population mean, .
c = 0:90, x = 58:62, s = 1:84, and n = 65.
Use the confidence interval, (36:482; 41:352), to answer questions 5 and 6.
5. What is the margin of error, E?
The critical value for 90% confidence interval is 1.645.
Margin of error E=1.645*1.84/sqrt(65)=0.375
6. What is the sample mean, x?
Sample mean is 58.62.
7. Let c = 0:99, s = 3:7, and E = 2:2. What is the minimum sample size, n, needed to estimate
? (Assume that s had been found from a preliminary sample that had more than 30 members.)
The critical value for 99% confidence interval is 2.576.
Margin of error=critical value*standard deivation/sqrt(n)
2.2=2.576*3.7/sqrt(n)
N=(2.576*3.7/2.2)^2=18.8≈19.
8. Repair Costs: Refrigerators In a random sample of 87 refrigerators, the mean repair cost
was $195:00 and the standard deviation was $25:50. Construct a 95% confidence interval for
the population mean repair cost.
The critical value for 95% confidence interval is 1.96
Margin of error=1.96*25.50/sqrt(87)=5.36
Upper limit: 195.00+5.36=200.36
Lower limit: 195.00-5.36=189.64
Therefore, the 95% confidence interval is [$189.64, $200.36].
9. Let c = 0:98 and n = 22. Find the critical value, tc.
Tc=TINV(0.02, 21) (0.02 is the probability, 21 is the degree of freedom, TINV is a function in excel)=2.518
Therefore, the critical t values are ±2.518.
10. Let c = 0:80, s = 5, and n = 14. Find the margin of error, E.
Use the information below to answer questions 11-13. Round your results to 3 decimal
places for this problem.
Let c = 0:90, x = 48:51, s = 2:15, and n = 19.
The critical t value for 80% confidence interval is 1.350
Margin of error=1.350*5/sqrt(14)=1.804
11. Construct a confidence interval for the population mean using the t-distribution.
The critical t value for 90% confidence interval is 1.734.
Margin of error=1.734*2.15/sqrt(19)=0.855
Upper limit: 48.51+0.855=49.365
Lower limit: 48.51-0.855=47.655.
Therefore, the 90% confidence interval is [47.655, 49.365].
12. Construct the confidence interval for the population mean that you would get if you incorrectly used a normal distribution.
The critical value for 90% confidence interval is 1.645.
Margin of error=1.645*2.15/sqrt(19)=0.811
Upper limit: 48.51+0.811=49.321
Lower limit: 48.51-0.811=47.699
Therefore, the 90% confidence interval is [47.699,49.321].
13. Which interval is wider?
The t interval is wider.
Use the information below to answer questions 14 and 15.
Repair Costs: Computers In a random sample of 12 computers, the mean repair cost was
$149:50 and the standard deviation was $37:89. Assume the repair cost is normally distributed.
14. Use a t-distribution to construct a 98% confidence interval for the population mean .
The critical value for 98% confidence interval is 2.718.
Margin of error=2.718*37.89/sqrt(12)=29.729
Upper limit: 149.50+29.729= 179.229
Lower limit: 149.50-29.729= 119.771
Therefore, 98% confidence interval is [$119.771, $197.299].
15. What is the margin of error, E, of ?
Margin of error=2.718*37.89/sqrt(12) =29.729
Use the table below to answer questions 16-18.
Incomes The monthly incomes for 14 randomly selected people, each with a bachelor's degree. Assume that the population of the monthly incomes is normally distributed.
3885.90 4533.40 6986.70 5100.30 7028.10 7915.40 5076.60
7150.40 5643.50 4628.88 3737.68 6030.58 7968.10 6372.27
16. What is the sample mean, x?
Sample mean ...

Solution Summary

The solution determines the margin of error calculations.

$2.19