STA 6166 UNIT 2 Section 1 Exercises
Welcome < Begin <

Section 1

< Unit 2 Section 1 Exercises > Section 2

Unit 2 Section 1 Exercises

You can choose to work some or all of the problems listed below. We recommend that you at least work the problems listed in your major area of interest. Answers to these exercises will be posted here (ANSWERS).

General Questions.
  1. How should you interpret the statement "the 95% confidence interval for the mean is (8.9, 12.7)"?
  2. As the level of confidence (a/2) goes up, does the sample size go up or down?
  3. As the underlying level of variance goes up, does the sample size go up or down?
  4. What are the five steps of a statistical test? (commit this to memory!)
  5. Define what is meant by a Type I error. (Commit this to memory also!)
  6. Define what is meant by a Type II error.
  7. Define what is meant by the Power of a test.
  8. With a two-tailed test, is the rejection region composed of one region or two?
  9. What does an OC curve tell us?
  10. What is a p-value? (see section 5.6).
  11. What is the differences between the z-test and the t-test?
  12. The shape of the t-distribution is indexed by one parameter. What is this parameter referred to as?
  13. Using the table of percentage points for the Student's t distribution given on page 1093, compute the following. ( n is the degrees of freedom).
    1. The value of t where P(T>t | n=26)=.05 (i.e. the 5% critical value for a t-distribution with 26 degrees of freedom.)
    2. P(T>3.169 | n=10) = ?
For students in agriculture and environmental fields.
  1. A oil company advertises that its new brand of tractor fuel significantly decreases fuel consumption in heavy use. You have been charged with determining whether the company's claim is valid. Your records indicated that the average fuel consumption for a standard configuration of tractor/plow is 10 liters per hour under a common standard plowing protocol (i.e. specified speed, depth of plow and soil conditions). You wish to determine if, using the new fuel, average fuel use is less than 10 l/hr. You decide to test this hypothesis by using a standard tractor/plow configuration and repeated measurements over similar field conditions. Initially you will do 15 of these replications, three per day for five days. At the end of the week, you will be testing a null hypothesis that the average fuel consumption is 10 l/h against an alternative that consumption is less than 10 l/h. Assuming the true standard deviation in consumption is roughly (s=) 1.5 l/h, and the test is being done at a Type I error rate of (a=) 0.05, what would the statistical power be for significant differences of +.5, +1.0 and +1.5 l/h (D) less than the expected 10 l/h.
  2. You decide to run the experiment with 15 replications, with the results produced in the following table. With these data, test the null hypothesis that nicotine content is 10 l/d against the alternative that the average is less than 10 l/d. (Note: The data table below can be directly cut and pasted into a spreadsheet or stat package - i.e. no html codes are imbedded.)
  3.  Replication Consumption
     1  8.99
     2  9.16
     3 10.20
     4  9.82
     5 11.54
     6  9.48
     7 10.65
     8  9.71
     9 11.43
    10  9.97
    11  9.11
    12  8.24
    13  8.44
    14  8.50
    15 11.34
     
  4. You are encouraged by the number of replicates with consumption less than 9.7 l/d and think that the true average consumption might really be 9.7 l/d. You want to do more sampling to get more definitive results. How many more replicates would you need to determine that the true population mean was really 9.7 l/d (the specific alternative hypothesis) instead of 10 l/d (the null hypothesis) if the true standard deviation is (s =) 1.0 l/d. The test is to be performed at a Type I error probability of (a=) 0.05 and we wish 85% power for the test (i.e. Type II error probability of b=0.15). [Hint: page 221]
  5. For comparison purposes, redo the test in step 2 using the Sign Test. How do the results compare?
For students in engineering fields.
  1. A copper mine manager is interested in installing new equipment that is expected to increase production. In the current process, average daily production varies by plus or minus 50 tons per day with an average of about 50 tons per day. The manager can perform a trial run with the new equipment for (n=) 15 days. At the end he will be testing a null hypothesis that the average is 50 t/d against an alternative that production is greater than 50 t/d. Assuming the true standard deviation of the process is roughly (s=) 100/4=25 t/d, and the test were being done at a Type I error rate of (a=) 0.05, what would the power be for significant differences of +10, +20 and +40 t/d (D) greater than the expected 50 t/d.
  2. The manager decided to run the experiment over 15-days with the results produced in the following table. With these data, test the null hypothesis that the average production is 50 tons per day against the alternative that the average is greater than 50 tons. (Note: The data table below can be directly cut and pasted into a spreadsheet or stat package - i.e. no html codes are imbedded.)
  3.  Day Yield
     1    57.8
     2    58.3
     3    50.3
     4    38.5
     5    47.9
     6   157.0
     7    38.6
     8   140.2
     9    39.3
    10   138.7
    11    49.2
    12   139.7
    13    48.3
    14    59.2
    15    49.7
     
  4. The mine manager is encouraged by the four days where production was over 70 and thinks that the true production mean might really be 70 t/d and wants to continue the trial to get more definitive results. How many days would be required to determine that the true population mean was really 70 t/d (the specific alternative hypothesis) instead of 50 t/d (the null hypothesis) if the true standard deviation is (s =) 40 t/d. The test is to be performed at a Type I error probability of (a=) 0.05 and we wish 80% power for the test (i.e. Type II error probability of b=0.20). [Hint: page 221]
  5. For comparison purposes, redo the test in step 2 using the Sign Test. How do the results compare?
For students in toxicology and health science fields.
  1. A tobacco company advertises that the average nicotine content of its cigarettes is at most 14 milligrams. You have been charged with determining whether the company's claim is valid. If the average nicotine is greater than 14.75 milligrams the company could be in trouble with government oversight agencies. You decide to test 20 cigarettes selected at random, one per day over 20 days of production. At the analysis step, you will be testing a null hypothesis that the average is 14 mg against an alternative that nicotine content is greater than 14 mg. Assuming the true standard deviation in nicotine content is roughly (s=) .8 mg, and the test were being done at a Type I error rate of (a=) 0.05, what would the power be for significant differences of +.2, +.4 and +.75 mg (D) greater than the expected 14 mg.
  2. You decide to run the experiment on 25 cigarettes, with the results produced in the following table. With these data, test the null hypothesis that nicotine content is 14 mg against the alternative that the average is greater than 14 mg. (Note: The data table below can be directly cut and pasted into a spreadsheet or stat package - i.e. no html codes are imbedded.)
     Cigarette Nicotine
     1 13.96
     2 15.33
     3 15.01
     4 14.46
     5 14.71
     6 15.35
     7 16.03
     8 15.22
     9 14.35
    10 15.56
    11 15.87
    12 14.82
    13 13.82
    14 14.19
    15 13.84
    16 14.33
    17 13.79
    18 15.32
    19 12.50
    20 14.17
    21 14.07
    22 14.58
    23 13.28
    24 16.60
    25 13.37
     
  3. You are concerned by the number of cigarettes with nicotine levels over 14.5 and think that the true production mean might really be 14.5 mg. You want to do more sampling to get more definitive results. How many more cigarettes would you need to determine that the true population mean was really 14.5 mg (the specific alternative hypothesis) instead of 14 mg (the null hypothesis) if the true standard deviation is (s =) .90 mg. The test is to be performed at a Type I error probability of (a=) 0.05 and we wish 90% power for the test (i.e. Type II error probability of b=0.10). [Hint: page 221]
  4. For comparison purposes, redo the test in step 2 using the Sign Test. How do the results compare?
For students in community development, education and social services fields.
  1. The University is concerned that married graduate students may be spending a large proportion of their graduate stipends on health care costs. When figuring out the total amount of stipends to offer, University officials factor in about $500 per year for health care expenses. You have been charged with determining, using a sample survey, if this number is a good estimate for average health care costs. If the average cost greater than $500, the University might increase stipends appropriately. As a first step, you decide to question 40 married male graduate students in the College of Agriculture, selected at random from the spring enrollment list. At the analysis step, you will be testing a null hypothesis that the average health care cost is $500 against an alternative that costs are greater than $500. Assuming the true standard deviation in costs is roughly (s=) $150, and the test is being done at a Type I error rate of (a=) 0.05, what would the power be for significant differences of +$50, +$75 and +$100 mg (D) greater than the expected $500.
  2. You decide to run the study with 40 students, with the results produced in the following table. With these data, test the null hypothesis that average costs is $500 against the alternative that the average is greater than $500. (Note: The data table below can be directly cut and pasted into a spreadsheet or stat package - i.e. no html codes are imbedded.)
    
     Person Expenditure
     1 417
     2 788
     3 605
     4 475
     5 577
     6 496
     7 585
     8 512
     9 482
    10 569
    11 505
    12 516
    13 408
    14 696
    15 575
    16 702
    17 596
    18 627
    19 444
    20 500
    21 534
    22 181
    23 338
    24 604
    25 519
    26 606
    27 482
    28 551
    29 550
    30 494
    31 356
    32 551
    33 426
    34 541
    35 511
    36 758
    37 328
    38 438
    39 520
    40 480 
     
  3. These data seem to indicate that true health care costs might be closer to $550 than to $500. You want to do more sampling to get more definitive results. How many more students would you need to determine that the true population mean was really $550 (the specific alternative hypothesis) instead of $500 (the null hypothesis) if the true standard deviation is (s =) $110. The test is to be performed at a Type I error probability of (a=) 0.05 and we wish 90% power for the test (i.e. Type II error probability of b=0.10). [Hint: page 221]
  4. For comparison purposes, redo the test in step 2 using the Sign Test. How do the results compare