O2. The following printout concerns a best subsets regression. Questions follow.

Transcription

1 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions O1. Frank Tanner is the lab manager at BioVigor, a firm that runs studies for agricultural food supplements. He has been asked to design a protocol for testing Porcine Plus, a diet additive that will improve weight growth in piglets from age six weeks to 4 weeks. The piglets to be used are Minnesota-Spotted, a variety for which the typical weight growth in this period is about 60 pounds, with a standard deviation of 1 pounds. How many piglets should Frank request if he d like his sample average to be within pounds of the true average with a probability of at least 80%? zα/ σ SOLUTION: This is governed by the formula n E. Frank should use E = (target error limit), σ = 1 (plausible standard deviation), z α/ = 1.8 (corresponding to 1 α = 0.80 and α/ = 0.10). This gets to n = So it looks like Frank should ask for 59 piglets. It s possible that Frank may want to be a little more conservative about the assessed standard deviation. If this product really does increase the weight gain, it would likely also increase the standard deviation. Using σ = 14, say, would lead to a request for 81 piglets. His managers might not want to supply all those piglets. On the other hand, they all end up as bacon and pork chops... O. The following printout concerns a best subsets regression. Questions follow. esponse is SWAMP 48 cases used W I L W H S D A U U L N I D T V F E E E Vars -Sq -Sq() C-p S E N Y X X X X X X X X X X X X X X X X (a) Give the names of the independent variables. (b) What is the dependent variable? (c) If you had to select the best three-predictor model, which predictors would you use? (d) Which is the best set of predictors to use? Indicate your reasons. 0 1 gs 01

2 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions SOLUTIONS: (a) There are four independent variables. These are WILDLIFE, WADEN, HUNTE, and SUVEY. (b) The dependent variable is SWAMP. (c) The best three-predictor model uses WILDLIFE, WADEN, and SUVEY. (d) The best model, using the C p criterion, clearly uses WILDLIFE and SUVEY. If you focus on or on s ε, you should reach the same conclusion. O3. At the end of each day, the unsold bread at rinsky s Bakery is weighed. The employees take some of it home, and the rest is donated to a homeless shelter. The daily amounts seem to represent a sample from a population with mean 3 pounds and standard deviation 1 pounds. The manager watches to make sure these unsold amounts stay under control. In particular, she ll audit operations if the amounts get too large. For the 31 days of October, the average daily amount was 36. pounds. What is the probability that, by chance alone, the average of a 31-day period would be 36. pounds or more? SOLUTION: Let X 1, X,, X 31 be random variables representing the daily unsold amounts. We should assume that these values are independent, each from a population with mean µ = 3 and σ = 1. It s not critical to assume that the population values follow a normal distribution. The sample average X will then have a distribution which 1 is approximately normal with mean 3 and with standard deviation It 31 follows then that X P[ X 36. ] = P P[ Z ] P[ Z 1.95 ] = 0.50 P[ 0 Z < 1.95 ] = = You could get a slightly more refined answer from software (or from a calculator that has a cumulative normal function); this would be This is a small probability, so you d have to say that the waste during October was unusually high. 0 gs 01

3 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions O4. The regression of W on G gave this Minitab output: egression Analysis: W versus G The regression equation is W = G Predictor Coef SE Coef T P Constant G S = Sq = 33.9% -Sq() = 3.4% Analysis of Variance Source DF SS MS F P egression esidual Error Total Please answer T (true) or F (false) to each of the following. (a) It is somewhat surprising that <. (b) A new data point with G new = 10 would lead to a prediction of W new = = 89. (c) The fit would be appraised as poor, because SS esidual Error > SS egression. (d) If all the data values (meaning all the G i s and all the W i s) were doubled, then the F statistic would be (e) If all the data values (meaning all the G i s and all the W i s) were doubled, the slope -10. would remain unchanged. (f) The regression slope would be regarded as not statistically significant. (g) The total sum of squares, namely 109,639, provides solid evidence of the effect of regression to mediocrity. (h) The correlation between the variables G and W is negative. (i) The data provide convincing evidence that increasing G by one unit will cause a decrease in W of 10. units. (j) Most of the residuals are between -50 and +50. SOLUTION: (a) is false. (b) is true. (c) is false. (d) is false. (e) is true. (f) is false. (g) is false. There s no surprise. This is what usually happens. That s exactly how a regression is to be used. The relative sizes of these, as measured through the F statistic, is what really matters. The F statistic would not change. The slope would not change. The t statistic for the regression is -4.70, which is easily significant. The two issues here are completely unrelated. (h), (i), and (j) solutions on next page. 0 3 gs 01

4 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions (h) is true. (i) is false. (j) is true. The correlation coefficient has the same sign as b 1, the slope estimate. Since b 1 = , we know that the correlation coefficient must also be negative. There is absolutely nothing here to suggest causation. The residuals, as a set of numeric values, have mean zero and standard deviation s ε = Thus about 3 of the residuals will be in (-41.05, 41.05). It follows that considerably more than 3 of the residuals are in (-50, 50). O5. For each of the following, please respond either impossible or could happen. You may use abbreviations I and CH. As illustrations, In a sample of size 15, the correlation between X and Y was 1.. The sample x 1, x,, x n produced an average x = impossible could happen (a) (b) In a multiple regression with four predictors, it was found that SS egression < SS esidual Error. In a multiple regression with four predictors, it was found that >. (c) Variables H and W have a correlation r HW = 0.48, and the linear regression of H on W produced a slope of (d) In the regression of W on Z, the fitted regression line passed through the point ( Z, W ). (e) The regression of Y on {G, H, J} had an value that is less than the value from the regression of Y on {G, H}. (f) In a simple linear regression of Y on X, the value of s ε can be a larger number than the standard deviation of the independent variable X. (g) In a very strong multiple regression, the F statistic was found as F = 1,810. (h) In a linear regression with n = 3, all the residuals were negative. (i) In a set of n = 59 values, the correlation was r X, Y = 0.9. Steve did the regression of Y on X and got the slope b 1 (Y on X) = 0.5, while Angela did the regression of X on Y and got the slope b 1 (X on Y) = (j) In a regression of Y on X, the slope was b 1 = 0.8, the correlation was r = 0., and the standard deviation of the x-values was 1,80. SOLUTION: (a) Could happen. The statement is equivalent to < (b) Impossible. It always happens that <. This can be seen in the formula (c) n 1 = 1 ( 1 ) which shows that 1 < 1 n 1. Just write as ( ). Impossible. The slope and correlation must have the same sign. 1 = 1, 0 4 gs 01

5 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions (d) Could happen. In fact, it has to happen as a simple consequence of b 0 = W - b 1 Z. (e) Impossible. As the set of predictors is enlarged, can only increase. (f) Could happen. Of course s ε and s X can only be compared in those cases in which Y and X are in the same units. (g) Could happen. You will actually see monster values for F now and then. (h) Impossible. The residuals must sum to zero. (i) Impossible. Observe that b 1 (Y on X) = Sxy Sxy and b 1 (X on Y) = S S. These (j) are not reciprocals of each other, except in the very special case = But having = 1.00 means that either r = +1 or r = -1, which is not the story here. Could happen. There are no contradictions in these statements. xx yy O6. The output below, based on a sample of used cars, shows the regression of price on age. Commas were inserted in large numbers to improve readability. Several of the positions are blank. Please supply the missing values. egression Analysis: Price versus Age The regression equation is Price = Age (a) Predictor Coef SE Coef T P Constant Age (b) S = -Sq = -Sq() = (c) (d) (e) Analysis of Variance Source DF SS MS F P egression 1,490,336,409,490,336, (f) esidual Error 171 3,644,398 (g) Total 3,113,58,549 (h) 0 5 gs 01

6 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions SOLUTION: (a) You can copy the intercept from the listing below. It s 16, Coef (b) Since t = SE Coef, this is 1, (c) The value in this position is s ε and it s recovered as MS esidual = 3,644,398 (d) (e) 1, The value is found as The SS egression SS Total value can be found as either =,490,336,409 3,113,58,549 1 s ε s Y To use the first formula, you ll need s Y = %. n 1 or as 1 ( 1 ) SSTotal n 1 = 3,113,58, s ε 1, , Then 1 = 1 s Y 4, = 79.87%. To use the second formula, note that = 1 in a simple regression. This 17 gives 1 ( ) MSegression (f) Find F = =,490,336, Since this is a one-predictor MS 3, 644,398 esidual regression, you can also get this as t = (-6.14) (g) This is obtained directly as the difference SS Total SS egression = 3,113,58,549,490,336,409 = 63,19,140. You could also get this from the relation SSesidual MS esidual =. Thus SS esidual = 3,644, ,19,058. (h) This is 17. The complete listing follows. Some of the results differ slightly because values were reconstructed from rounded information.. egression Analysis: Price versus Age The regression equation is Price = Age Predictor Coef SE Coef T P Constant Age S = Sq = 80.0% -Sq() = 79.9% 0 6 gs 01

7 STAT-UB.0103 Exam 01.APIL.11 OVAL Version Solutions Analysis of Variance Source DF SS MS F P egression 1,490,336,409,490,336, esidual Error ,19,141 3,644,398 Total 17 3,113,58,549 O7. The reserve fund at public radio station WXXQ currently sits at $4,000. Today is the first day of their annual fund drive. Based on past experience, they believe that the daily changes in the reserves, based on random contributions and random expenses, will have a mean of $3,00 and a standard deviation of $,100. The station would like to know the probability that the reserve will reach or exceed $80,000 in fourteen days. (a) [5 points] Please state any assumptions that are helpful in doing this work. (b)[15 points] Find the actual probability, using the numbers above and your assumptions in (a), that the reserve will reach or exceed $80,000 in fourteen days. SOLUTION: For (a) there are two critical assumptions. You ll need to assume that the daily values are independent of each other. You ll also need to assume that the daily changes come from a normal population. The sample of n = 14 is not officially enough to invoke the Central Limit theorem. For (b), let T be the total for 14 days. Let 0 = $4,000 be the current reserve, and let 14 be the reserve after 14 days. Observe that 14 = 0 + T. The mean of the distribution of T is 14 $3,00 = $44,800, and the standard deviation of the distribution of T is $, $7, Then P[ 14 $80,000 ] = P[ 0 + T $80,000 ] = P[ T $38,000 ] = T $44,800 $38, 000 $44,800 P $7, $7, P[ Z ] = P[ 0 Z ] P[ 0 Z 0.87 ] = = Software will get the slightly more precise answer The station has a very good chance of reaching the $80,000 objective. 0 7 gs 01