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Exercise 12
The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds.
1. What is the value of the population mean? What is the best estimate of this value?
2. Explain why we need to use the t distribution. What assumption do you need to make?
3. For a 90 percent confidence interval, what is the value of t?
4. Develop the 90 percent confidence interval for the population mean.
5. Would it be reasonable to conclude that the population mean is 63 pounds?
Exercise 28
A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. 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 0.2 pounds?
Exercise 6
The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The quality-assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?
Exercise 18
The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?
Exercise 24
Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the afternoon shift than on the day shift. A sample of 54 day-shift workers showed that the mean number of units produced was 345, with a standard deviation of 21. A sample of 60 afternoon-shift workers showed that the mean number of units produced was 351, with a standard deviation of 28 units. At the .05 significance level, is the number of units produced on the afternoon shift larger?
Exercise 38
Two boats, the Prada (Italy) and the Oracle (U.S.A.), are competing for a spot in the upcoming America's Cup race. They race over a part of the course several times. Below are the sample times in minutes. At the .05 significance level, can we conclude that there is a difference in their mean times?
Boat Times (minutes)
Prada (Italy) 12.9 12.5 11.0 13.3 11.2 11.4 11.6 12.3 14.2 11.3
Oracle (U.S.A.) 14.1 14.1 14.2 17.4 15.8 16.7 16.1 13.3 13.4 13.6 10.8 19.0
Exercise 30
There are four auto body shops in a community and all claim to promptly serve customers. To check if there is any difference in service, customers are randomly selected from each repair shop and their waiting times in days are recorded. The output from a statistical software package is:
Summary
Groups Count Sum Average Variance
Body Shop A 3 15.4 5.133333 0.323333
Body Shop B 4 32 8 1.433333
Body Shop C 5 25.2 5.04 0.748
Body Shop D 4 25.9 6.475 0.595833
Exercise 12
For many years TV executives used the guideline that 30 percent of the audiences were watching each of the prime-time networks and 10 percent were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the .05 significance level, can we conclude that the guideline is still reasonable?
Exercise 26
A study regarding the relationship between age and the amount of pressure sales personnel feel in relation to their jobs revealed the following sample information. At the .01 significance level, is there a relationship between job pressure and age?
Degree of Job Pressure
Age (years) Low Medium High
Less than 25 20 18 22
25 up to 40 50 46 44
40 up to 60 58 63 59
60 and older 34 43 43
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This problem set includes 9 questions. All answers and complete, step-by-step explanations are included in the solution.
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a. Lind Chapter 9
Exercises 12,
The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds.
1. What is the value of the population mean? What is the best estimate of this value?
2. Explain why we need to use the t distribution. What assumption do you need to make?
3. For a 90 percent confidence interval, what is the value of t?
4. Develop the 90 percent confidence interval for the population mean.
5. Would it be reasonable to conclude that the population mean is 63 pounds?
We don't know the population mean, but to estimate it, we can look at the mean of a sample of that population. The sample mean is 60 lbs, but we don't know how accurate that is in predicting the population mean until we use a t-test.
We need to use the t-distribution instead of the z-distribution because we know the sample standard deviation, not the population standard deviation. To use a t-test you have to assume that the values are normally distributed. If you think they're not, you need to use another statistical test.
To find the confidence interval, we need to find the critical value for t using a distribution table. We need to find the table with the correct degrees of freedom, which is calculated as df = n - 1. Here, that number is df = 16 - 1 = 15. (You can find distribution tables here: http://www.statsoft.com/textbook/sttable.html#t)
The 90% critical value for a two sided test is t = 1.753.
To calculate the 90% confidence interval, we use the following formula:
where y is the sample mean (60), t is the critical value we want (1.753), s is the sample standard deviation (20), and n is the sample size (16).
So, our confidence interval is:
60 ± 1.753(20/√16)
60 ± 1.753(5)
60 ± 8.765
Or, the confidence interval is (51.235, 68.765). That means there is a 90% chance that the true population means falls in this interval. Therefore it is reasonable to assume that 63 lbs is a good estimate for the population mean, although it is unlikely that the true population mean is exactly 63 lbs.
Exercise 28
A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. 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 0.2 pounds?
So, to summarize, sample mean = 20.4, sample standard deviation = 0.5, and the sample size = 20.
Remember how we did the confidence interval in the ...
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