STAT-UB.0103 Spring 2012 Homework Set 8 Solutions

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1 1. Curtis was investigating the properties of a new vacuum cleaner motor in terms of a standardized lab procedure. In this procedure, the motor was applied to a pile of gypsum powder for five seconds, and the weight sucked into the vacuum bag was measured. The operational standard, the average from the previous design, was 220 grams. Curtis set out to test H 0 : µ = 220 versus H 1 : µ 220 with a set of 20 trials. This set of trials produced an average of 234 grams, with a standard deviation of 44 grams. (a) Identify the symbols n, µ, σ, x, s. (b) At the 5% level of significance does Curtis accept H 0 or reject H 0? SOLUTION: (a) Here µ is the unknown mean of the population of measurements for the new motor. Also, σ is the unknown standard deviation of that population. Of course x = 234 is the sample average and s = 44 is the sample standard deviation. The sample size is n = 20. The value 220 is the comparison, and in most notational schemes, it s denoted as µ 0. (b) You can do this in steps, if desired. Step 1: H 0 : µ = 220 vs H 1 : µ 220 Step 2: α = 0.05 X µ 0 Step 3: The test statistic is t = n, where n = 20 and µ0 = 220. The s degrees of freedom number for t is n 1 = 19. The null hypothesis H 0 will be rejected if t t 0.025; 19 = Step 4: Find t = Step 5: Since = < 2.093, H 0 must be accepted. This motor beat the previous values of 220 g, but the improvement was not big enough to be declared statistically significant. If you use Minitab s Stat Basic Statistics 1-Sample t Summarized data, fill out the information panel to look like this: 1

2 The result is One-Sample T Test of mu = 220 vs not = 220 N Mean StDev SE Mean 95% CI T P (213.41, ) Stop by our store with this coupon for free coffee and croissants and check out our new six-foot HDTV screens. An advertising flyer with this line was distributed to people on the street, and it was anticipated that 10% of the people receiving the flyer would stop at the store. From 590 flyers given out on the street, there were 48 who came to the store. At the 5% level of significance, does this differ from the anticipated 10%? (a) Identify the symbols n, α, p, p, p 0. (b) Perform the requested test. SOLUTION: (a) Here n is the sample size 590 α is the probability of type I error 0.05 p is the event proportion in the data p is the true-but-unknown population parameter. p 0 = 0.10 is the comparison value. (b) Here are the steps to the hypothesis test. Step 1: H 0 : p = 0.10 vs H 1 : p 0.10 Step 2: α = 0.05 pˆ p0 Step 3: The test statistic is Z = n p 1 p ( ) 0 0, where n = 590 and p 0 = The null hypothesis H 0 will be rejected if Z z = Step 4: Find p = Then Z = Step 5: Since < 1.96, H 0 must be accepted The flyer had a sample yield of about 8.1%. This is below the anticipated 10%, but not significantly so. 2

3 You can do this with Minitab, but be careful. Do Stat Basic Statistics 1 Proportion. Set the Test Proportion to Also do Options and check off the box Use test and interval based on normal approximation. The panel will be this: The results will be these: Test and CI for One Proportion Test of p = 0.1 vs p not = 0.1 Sample X N Sample p 95% CI Z-Value P-Value ( , ) Using the normal approximation. 3. Suppose that you have these data from populations that are assumed normal: Sample n Average St. Dev You would like a 95% confidence interval for the parameter difference µ 1 - µ 2. Please give this interval * when you allow unequal population standard deviations * when you assume that the population standard deviations are equal For each of these two setups, note also the results of testing H 0 : µ 1 = µ 2 versus H 1 : µ 1 µ 2 at the 5% level. You can do this by hand, but Minitab makes it so much easier! SOLUTION: Should we do the work with assumed equal population standard deviations? The sample standard deviations have the ratio , so the decision is marginal. 3

4 Suppose that you allow unequal standard deviations. Use Minitab s Stat Basic Statistics 2 Sample t. Fill in the resulting panel as follows: The result is this: Two-Sample T-Test and CI Sample N Mean StDev SE Mean Difference = mu (1) - mu (2) Estimate for difference: % CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 39 If you had asked for equal standard deviations, you would have checked off the box Assume equal variances. This would give Two-Sample T-Test and CI Sample N Mean StDev SE Mean Difference = mu (1) - mu (2) Estimate for difference: % CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 48 Both use Pooled StDev = The two versions of the confidence interval are very close. The test statistics are and -0.81, also close. 4

5 4. The Excelsior Mutual Fund Family specializes in attentive client service and vigilant market surveillance. At least, it says this in its advertising, in the hopes of justifying its high service charges and 8% front-end load. The fund currently has many thousands of clients, and it has undertaken a survey of these clients, obtaining from a random sample of 320 of them an average age of 56.4 years, with a standard deviation of 8.1 years. (a) Give a 95% confidence interval for the average age of the population of clients. (b) Give a 99% confidence interval for the average age of the population clients. (c) It has been noted that the population of ages seems not to follow a normal distribution. In particular, the data seem to be slightly negatively skewed. How does this observation influence the procedures used in steps (a) and (b)? SOLUTION: For part (a), the confidence interval is x ± t 0.025; 319 sn, which is 56.4 ± or 56.4 ± The 319 degrees of freedom certainly permits us to use the line of the t table; equivalently we could obtain 1.96 as the two-sided 5% point from the normal distribution. Minitab computes the t value more precisely as t 0.025; 319 = You are 95% confident that the true mean lies in the interval (55.51, 57.29). You could round this to (55.5, 57.3), since 0.01 year is about three days, and it s hard to imagine that the users of this information are interested in that kind of precision. For (b) the procedure is the same as in part (a), except that the two-sided 1% point from the normal distribution, 2.58, replaces the value The resulting interval is then ± 2.58, or 56.4 ± Minitab would give t 0.005; 319 = The 320 interval can be given as (55.23, 57.57), which could certainly be rounded to (55.2, 57.6). This interval has length = 2.34, while the interval in (a) has length The ratio is that of the table points which were used, namely For this example, the 99% confidence interval is about 32% longer. With regard to (c), with a sample size of 320, there is no need to worry about the distribution from which the values were obtained. 5. The distributor of a certain variety of tomato seed has promised that 80% of the seeds will germinate under standard greenhouse conditions. You test this claim with 500 seeds, and you find that 362 germinate successfully. At the 5% level of significance, test the claim as H 0 : p = 0.80 against the alternative H 1 : p

6 SOLUTION: Here are the steps: Step 1: H 0 : p = 0.80 vs H 1 : p 0.80 Step 2: α = 0.05 pˆ p0 Step 3: The test statistic is Z = n p 1 p ( ) 0 0, where n = 500 and p 0 = The null hypothesis H 0 will be rejected if Z z = Step 4: Find p = = Then Z = = Step 5: Since > 1.96, H 0 must be rejected. A good statistician would also check to be sure that the experiment was performed carefully. After all, seeds put up for test are usually all treated together in all small number of seed trays, and statistical independence could be badly compromised. Care is needed if you do the arithmetic through Minitab. Do Stat Basic Statistics 1 Proportion. Click on Options to set the Test Proportion to Also check off the box Use test and interval based on normal approximation. The results will be these: Test and CI for One Proportion Test of p = 0.8 vs p not = 0.8 Sample X N Sample p 95% CI Z-Value P-Value ( , ) If you do not check of the box for the normal approximation, the test is done in a completely different style. You won t get a test statistic, but you ll still get the p-value reported as The weekly salaries of union electricians in two metropolitan areas were compared by taking random samples of 50 in each area. The results were these: Area n Mean Standard deviation At the 5% level of significance, test the hypothesis H 0 : µ 1 = µ 2 versus the alternative H 1 : µ 1 µ 2. SOLUTION: Since s 1 and s 2 are within a factor of two of each other, it would be most reasonable to use the equal-standard deviation form of the t statistic. This assumes that σ 1 = σ 2. Here are the steps: Step 1: H 0 : µ 1 = µ 2 versus H 1 : µ 1 = µ 2. 6

7 Step 2: α = n1 n2 x1 x2 Step 3: The test statistic will be t = n + n s 1 2 p where s p is the pooled standard deviation s p = with 98 degrees of freedom, ( n 1) s + ( n 1) s n + n The null hypothesis will be rejected if t t 0.025;98 = The value of t 0.025;98 was obtained from Minitab. Step 4: From the given information, we have 2 (50 1) (50 1) s p = , , and then s p The t statistic is then t = Step 5: The null hypothesis H 0 must be rejected. x1 x2 If you had computed Z = to avoid the assumption of equal standard deviations, 2 2 s1 s2 + n1 n2 you would have produced the same test statistic to three significant figures! In Minitab, you can use Stat Basic Statistics 2 Sample t Summarized data. If you check the box Assume equal variances, you get the following:. Two-Sample T-Test and CI SE Sample N Mean StDev Mean Difference = mu (1) - mu (2) Estimate for difference: % CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 98 Both use Pooled StDev = x1 x2 The Z form of the test, namely s s + n n , has value as well. 7

8 7. The management at Jackson & Flinch brokerage services has established dollar quotas for each of their brokers, but the brokers are generally not informed about these quotas. As part of an experiment, 21 brokers (selected at random) were actually told their annual dollar quotas just to see how knowledge of the quota would influence their performance. These brokers were compared with 15 others who were not told their quotas. The data below indicate the fraction of annual quotas achieved by the end of October: Told about quota? n Mean Standard deviation No Yes At the 0.05 level of significance, test the null hypothesis H 0 : µ NO = µ YES against the alternative H 1 : µ NO µ YES. SOLUTION: Begin by assuming that the samples represent values from normal populations with equal standard deviations. The small sample sizes require the assumption of normal distributions, and certainly the computed standard deviations make it reasonable to claim that σ NO = σ YES. Then proceed as follows: Step 1: H 0 : µ NO = µ YES versus H 1 : µ NO = µ YES. Step 2: α = n1 n2 x1 x2 Step 3: The test statistic will be t = n + n s 1 2 p where s p is the pooled standard deviation s p = ( n 1) s + ( n 1) s n + n NO NO YES YES NO YES with 34 degrees of freedom,. The null hypothesis will be rejected if t t 0.025;34 = The value of t 0.025;34 was obtained from Minitab. Step 4: From the given information, we have (15 1) (21 1) s p = , and then s p The t statistic is then t = Step 5: The null hypothesis H 0 must be accepted. We have a clear case of nonsignificance. That is, we declare that there is no significant difference between the two sets of brokers. 8

9 Minitab can do this with Stat Basic Statistics 2-Sample t Assume equal variances. The result: Two-Sample T-Test and CI Sample N Mean StDev SE Mean Difference = mu (1) - mu (2) Estimate for difference: % CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 34 Both use Pooled StDev = As part of an investigative process, a financial reporter has been comparing the advice given by two investment newsletters. Each time a buy recommendation was given, he noted the stock price; he then recorded whether the price was higher in exactly three months. Each case in which the price was higher was called a success. Here is the summary of his findings: Newsletter Success Failure Total number of buy recommendations Third Millenium Global Starship Based on these values, does the reporter have an interesting story? SOLUTION: This question can be answered easily with the chi-squared test. Complete the table to show a total row: Newsletter Success Failure Total number of buy recommendations Third Millenium Global Starship Total Here are the hypothesis testing steps. Step 1. H 0 : p TM = p GS versus H 1 : p TM p GS where p TM and p GS are the success probabilities for the two newletters. Step 2. α = Step 3. The test statistic is χ 2, and the null hypothesis will be rejected when χ Step 4. Numerically, we find χ 2 123a f = Step 5. Certainly H 0 is accepted. There is not sufficient evidence to reject H 0. 9

10 The reporter would have to conclude that there are no significant differences between the newsletters with regard to these buy recommendations. The observed success rate for Third Millenium was %, and for Global Starship was = 55%. Minitab can be used here as well. Use Stat Basic Statistics 2 Proportions and fill the panel as follows: Click on Options and fill in as follows: The pooled estimate form of this test will give you a normal statistic (Z) with the property Z 2 = χ 2. In this case, the output is Test and CI for Two Proportions Sample X N Sample p Difference = p (1) - p (2) Estimate for difference: % CI for difference: ( , ) Test for difference = 0 (vs not = 0): Z = 1.09 P-Value = Fisher's exact test: P-Value = Observe that =

11 9. People on marketing panels are often given products to use at home and evaluate. (Plenty of people really enjoy being on these panels.) In one panel, 75 people were given a one-gallon jug of Super Cal-Plus Orange Juice to take home, and another 78 people were given a one-gallon jug of CitrusLove Orange Juice to take home. When the panelists returned exactly one week later they were asked, Did you finish the gallon of orange juice? Here are the responses: Juice Brand Finished Juice Did Not Finish Juice Total Super CitrusLove Total At the 0.05 level of significance, is there a difference between these two juices for this particular question? SOLUTION: We should interpret the is there a difference comment as asking for a hypothesis test. Here are the steps. Step 1. Step 2. α = Step 3. H 0 : p S = p C versus H 1 : p S p C where p S and p C are the probabilities of finishing the gallon for the two brands. The test statistic is χ 2, and the null hypothesis will be rejected when χ ( ) 2 Step 4. Numerically, we find χ = Step 5. Certainly H 0 is rejected. The probabilities for finishing the gallon jug are significantly different. Of course, you would certainly note that the probability of finishing the gallon jug are p S = % for Super Cal-Plus and p = 54 C 69% for CitrusLove. 78 If you use Minitab (as described in the previous problem) you will get Test and CI for Two Proportions Sample X N Sample p Difference = p (1) - p (2) Estimate for difference: % CI for difference: ( , ) Test for difference = 0 (vs not = 0): Z = P-Value = Fisher's exact test: P-Value = Of course these agree, as (-3.31) 2 =

12 10. MBS10, problem 7.34, page 434. The file is CRASH.MTW. The variables of interest are DRIVCHST (column C13) and PASSCHST (column C14). These data will also be used for the next two problems. SOLUTION: a. The parameter of interest is µ D - µ P, the difference between the mean driver chest injury and the mean passenger chest injury. If would be perfectly reasonable to define µ DIFF = µ D - µ P. b. For this experiment, 98 vehicles were used. A major factor is this experiment is certainly going to be the fact that not all the collisions were equally severe. Thus, the difference (driver injury) i (passenger injury) i represents the distinction between the two dummies both in the same collision. c. In Minitab, create a new column for the difference C13 C14. It should be understood that positive values mean that the driver had a more severe injury. You might call this new variable as DiffCHST. The differences are small integers, and you can see what they look like through Graph Dotplot. The display counts 46 negatives (passenger injury worse), 10 zeroes, and 42 positives (driver injury worse), and this suggests that the passenger dummies had worse chest injuries. It remains to be seen whether there is statistical significance here or just chance. This new variable has mean and standard deviation A 99% confidence interval for the true difference is ± t 0.005; We can use t 0.005; 97 = (from Minitab) or just use 2.660, the printed value for t 0.005; 60 in the text version of the t table. The interval is then ± This interval is (-2.025, 0.903). 12

13 We could also get the interval from Minitab. Do Stat Basic Statistics Paired t. Set the panel as follows: For this problem, you ll need to use Options to set the desired confidence level to The result is this: Paired T-Test and CI: DRIVCHST, PASSCHST Paired T for DRIVCHST - PASSCHST N Mean StDev SE Mean DRIVCHST PASSCHST Difference % CI for mean difference: ( , ) T-Test of mean difference = 0 (vs not = 0): T-Value = P-Value = This agrees with the numeric calculations above. Also, a test statistic is given, and this will be useful soon. d. It would seem that the passenger injury is slightly worse. The confidence interval overlaps zero, so we d have to say that the difference in injury level is not statistically significant. Ouch. The experimenters had to crash 98 vehicles to learn this. e. The statistical procedure requires independent crashes. This seems to be a credible assumption. It also requires normal distributions, but with n = 98, we are somewhat unconcerned about normal distributions. 11. This is a continuation of the previous problem. Provide a reasonable graphical display which investigates the comparison between the driver chest injury level and the passenger chest injury level. 13

14 SOLUTION: There are many pictures that show what s going on. Here is a simple scatterplot with driver injury on the vertical axis and passenger injury on the horizontal. 70 Scatterplot of DRIVCHST vs PASSCHST 60 DRIVCHST PASSCHST It would be helpful to insert the line on which DRIVCHST = PASSCHST. This can be done with the Minitab drawing palette. The result is this: 70 Scatterplot of DRIVCHST vs PASSCHST 60 DRIVCHST PASSCHST Points above the line have worse injuries to the driver, and points below the line have worse injuries to the passenger. 14

15 12. At the 5% level of significance, test the null hypothesis that the driver and passenger means are the same against the alternative that they are different. (Please note that this asks for a 5% test, whereas problem 9.32 from MBS9 worked with a 99% confidence interval.) SOLUTION: Step 1: H 0 : µ D - µ P = 0 versus H 1 : µ D - µ P 0 Step 2: α = 0.05 Step 3: The test statistic is t = This could also be phrase as H 0 : µ D = µ P versus H 1 : µ D µ P. If you define µ DIFF = µ D - µ P, then the problem could be given as H 0 : µ DIFF = 0 versus H 1 : µ DIFF 0. DIFF n s DIFF with n 1 = 97 degrees of freedom. Here H 0 will be rejected if t t 0.025; 97 = This cutoff was obtained from Minitab, but it would reasonable to use t 0.025; 60 = from the table in MBS9. Step 4: As noted previously, the mean difference is , with a standard deviation of The test statistic is t = This t value was given in the Minitab calculation shown for problem 9.32c of MBS9. The p-value was noted as Step 5: This t value is nowhere close to statistical significance. We will accept the null hypothesis that the passenger and driver injury levels are the same. Minitab will do this directly from the spreadsheet. The command is Stat Basic Statistics Paired t. The output was listed with problem 10, using MBS10, problem 7.34, page