Dr. Patton is a Professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal...

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Dr. Patton is a Professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 40 students, the mean number of misspelled words per essay was 6.05. Construct a 95 percent confidence interval for the mean number of misspelled words in the population of student essays.
12. Use Appendix A.2, Excel or MegaStat to locate the value of
t
under the following conditions.
a) The sample size is 15 and the level of confidence is 95 percent.
b) The sample size is 24 and the level of confidence is 98 percent.
c) The sample size is 12 and the level of confidence is 90 percent.
16. The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 27 kg with a standard deviation of 9 kg. Assume a normal population.
a) What is the value of the population mean? What is the best estimate of this value?
b) Explain why we need to use the
t
distribution. What assumption do you need to make?
c) For a 90 percent confidence interval, what is the value of
t?
d) Develop the 90 percent confidence interval for the population mean.
e) Would it be reasonable to conclude that the population mean is 28 kg?
24. Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for sporting events, proms, birthdays, and other special events. Zack Schadek, the owner, received a large shipment this morning. To ensure the quality of the shipment, he selected a random sample of 300 cups. He found 15 to be defective.
a) What is the estimated proportion defective in the population?
b) Develop a 95 percent confidence interval for the proportion defective.
c) Zack has an agreement with his supplier that he is to return lots that are 10 percent or more defective. Should he return this lot? Explain your decision.
36. A processor of carrots cuts the green top off each carrot, washes the carrot, and inserts six to a package. Twenty packages are inserted into a box for shipment. To test the mass of the boxes a few were checked. The mean mass was 9.3 kg, the standard deviation 0.23 kg. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.09 kg.
48. A recent study by an automobile dealer revealed the mean amount of profit per car sold for a sample 20 salespeople was $290, with a standard deviation of $125. Develop a 95 percent confidence interval for the population mean.
60. You need to estimate the mean number of travel days per year for outside salespeople. The mean of a small pilot study was 150 days, with s standard deviation of 14 days. If you must estimate the population mean within 2 days, how many outside salespeople should you sample? Use the 90 percent confidence level.
Answered Same DayDec 23, 2021

Answer To: Dr. Patton is a Professor of English. Recently she counted the number of misspelled words in a group...

Robert answered on Dec 23 2021
131 Votes
Complete the following textbook exercises:
10, 12, 16, 24, 36, 48 and 60
10. Dr. Patton is a Professor of English. Recently she c
ounted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 40 students, the mean number of misspelled words per essay was 6.05. Construct a 95 percent confidence interval for the mean number of misspelled words in the population of student essays.
The interval is 6.05 ± 1.96*2.44/√40 = 6.05± 0.7561704 = 5.2938 to 6.8062
12. Use Appendix A.2, Excel or MegaStat to locate the value of t under the following conditions.
ASSUMING 2 TAIL TEST
a) The sample size is 15 and the level of confidence is 95 percent.
Degree of freedom = 15-1=14
T value = 2.145
b) The sample size is 24 and the level of confidence is 98 percent.
Degree of freedom = 24-1=23
T value = 2.5
c) The sample size is 12 and the level of confidence is 90 percent.
Degree of freedom = 12-1=11
T value = 1.796
16. The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 27 kg with a standard deviation of 9 kg. Assume a normal population.
a)...
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