oriv. Tukey Confidence Interval252y0772 11/26/07 (Page layout view!)ECO252 QBA2 Name __Key_________________ THIRD EXAM Student number_______________ November 29, 2007 Class Day and hour____________ Version 2I. (8 points) Do all the following (2 points each unless noted otherwise). Make Diagrams! Show your work! All probabilities must be between zero and (positive) 1. 14,27~ Nx1. 7746 xP142775142746zP 43.336.1 zP 36.1057.30 zPzP0867.4131.4998. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between 1.36 and 3.43. Because this is on one side of zero we must subtract the area between zero and 1.36 from the larger area between zero and 3.43. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 27. Indicate the mean by a vertical line! Shade the area between 46 and 77. This area is on one side of the mean (27) so we subtract to get our answer. 2. 3921 xP142739142721zP 86.043.0 zP 86.00043.0 zPzP4715.3051.1664. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the entire area between -0.43 and 0.86. Because this is on both sides of zero we must add the area between -0.43 and zero to the area between zero and 0.86. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 27. Indicate the mean by a vertical line! Shade the entire area between 21 and 39. This area is on both sides of the mean (27) so we add to get our answer. 3. 0xP14270zP 93.1 zP 0093.1 zPzP9732.5.4732. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the entire area above -1.86. Because this is on both sides of zero we must add the area between -1.86 and zero to the area above zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 27. Indicate the mean by a vertical line! Shade the entire area above zero,remembering that zero is below the mean. This area is on both sides of the mean (27) so we add to get our answer. 4. 08.x (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve witha mean at 0. 08.z is the value of z with 8% of the distribution above it. Since 100 – 8 = 92, it is also the 92nd percentile. Since 50% of the standardized Normal distribution is below zero, your diagram shouldshow that the probability between 08.z and zero is 92% - 50% = 42% or .4200.008. zzP The closest we can come to this is .4203.41.10 zP(1.40 is also pretty close.) So41.108.z (or something slightly smaller). To get from 08.z to 08.x, use the formula zx , which is the opposite of xz . 74.461441.127 x. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 27. Show that 50% of the distribution is 1252y0772 11/26/07 (Page layout view!)below the mean (27). If 8% of the distribution is above 08.x, it must be above the mean and have 42% ofthe distribution between it and the mean.Check: 74.46xP142774.46zP 08.0793.4207.5.41.10041.1 zPzPzP2252y0772 11/26/07 (Page layout view!)II. (22+ points) Do all the following (2 points each unless noted otherwise). Do not answer a question ‘yes’ or ‘no’ without giving reasons. Show your work when appropriate. Use a 5% significance level except where indicated otherwise. Note that this is extremely long and that no one will do all the problems, so look them over!1. Turn in your computer problems 2 and 3 marked as requested in the Take-home. (5 points, 2 point penalty for not doing.)2. In an ordinary 1-way ANOVA, if the computed F statistic exceeds the value from the F table at the given significance level, we cana. Reject the null hypothesis because the difference between the means is not significantb.* Reject the null hypothesis because there is evidence of a significant difference between some of the means.c. Not reject the null hypothesis because the difference between the means is not significant.d. Not reject the null hypothesis because the difference between the means is significant.c. Not reject the null hypothesis because the difference between the variances is not significant.d. Not reject the null hypothesis because the difference between the variances is significant.e. None of the above. [7]3. After an analysis if variance, you would use the Tukey-Kramer procedure or similar confidence intervals to check a. For Normalityb. For equality of variancesc. For independence of error termsd.* For pairwise differences in meanse. For all of the abovef. For none of the above4. If an ordinary one-way ANOVA has 25 columns 17 rows and 4252517 , the degrees of freedom for the F test are a. 400 and 24b. 408 and 16c.* 24 and 400d. 16 and 408e. 400 and 424f. 408 and 424g. 424 and 400h. 424 and 408i. 16 and 24j. None of the above. The correct answer is _______.Explanation: This is a one-way ANOVA. The total number of observations is 425n and the number of columns is 25m. This means there are 425-1 = 424 total degrees of freedom and that between the columns there are 25-1 = 24 degrees of freedom. This leaves 424 – 24 = 400 degrees of freedom for the error (within) term. Numbers are filled in below.Source SS DF MS F FBetween SSB1m241mSSBMSBMSWMSBF 40.1400,2401.FWithinSSW400 mnmnSSWMSWTotal SST4241 n3252y0772 11/26/07 (Page layout view!)5. Assuming that your answer to 4 is correct and that the significance level is 1%, the correct value of F from the table is __1.84___. (This may have to be approximate. If so, what did you use?) (1)[12] Note: I will check your answer against what you said in the previous question. The answer above is wrong if you did not say something close to 400,2401.F.Exhibit 1 A manager believes that the number of sales that an employee makes is related to the number ofyears worked and their score on an aptitude test. He runs the
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