Managerial Statistics Practice Final Exams Solutions, Study notes of Probability and Statistics

Download Managerial Statistics Practice Final Exams Solutions and more Study notes Probability and Statistics in PDF only on Docsity! COLUMBIA BUSINESS SCHOOL Fall 2001 Professor Paul Glasserman B6014: Managerial Statistics 403 Uris Hall Solutions to Practice Final Exams #1 and #2 EXAM #1 1. a) Let µ be true average number of calories in the Diet-Burger. The null hypothesis, H0, is µ = 400 and the alternative is µ >400. Therefore, the probability of a Type I error (α) is just the probability of rejecting the null when it is true. If the null is true, then µ = 400, so we need to calculate (recall n = 40 and σ=30): .0174.0)11.2( 40/30 400410)410( =>=
   −>=> ZPZPXP The probability of a Type I error is only 1.74%. b) Answer: higher. With a sample of size 10, the likelihood of making a Type I error increases since the standard error increases and thus we are more likely to get a sample mean that deviates from the population mean by 10 units. c) Assume µ=422. Then the null hypothesis is FALSE. A Type II error occurs when a false null is not rejected (i.e., is accepted). This occurs when the sample mean is less than 410. So we must calculate (recall n = 40 and σ = 30): .0057.0)53.2( 40/30 422410)410( =−<=
   −<=< ZPZPXP The probability of a Type II error is only 0.57%. 2. This is a hypothesis test. Let p be the proportion of coupons that are redeemed. We set the null hypothesis to be that there has not been any change in people coupon usage habits. I.e., H0: p = 0.06 and HA: p ≠ 0.06. The sample proportion is p̂ = 315/5431 = 5.80%. Let's calculate the p-value. .2676.0)62.0( 5431 )06.01(06.0 06.0058.0)058.0ˆ( =−<=
      − −<=< ZPZPpP The p-value is double this number, so p-value=2(0.2676) = 53.52%. This is a rather high p-value that suggests that there is little evidence against the null hypothesis. (Put another way, if p=0.06 so that 6% of coupons are redeemed on average, then there is a 53.52% chance of observing a sample that deviates from 6% by more than the amount 2 we observed. So this suggests that a sample like the one we have is very likely to occur by chance, even if the population proportion is 6%.) The data do not suggest that the proportion of responses is different from 6%. 3. The sample proportion of failures is 12/50 = 0.24. The 99% confidence interval is: 0.24 ± 2.58 50 )24.01)(24.0( − = 0.24 ± 0.156 = (0.084,0.396). 4. a) (A) t-stat for TVAD: 0.127/0.017=7.47 (B) p-value for TVAD: almost 0% (C) R-squared: SSR/SST=(1.971)/(1.971+0.447)=0.815. (D) Standard Error of Estimate: se = 1−− Kn SSE = 18 447.0 = 0.1575. (E) Degrees of Freedom, Error: n−K−1=n−3−1=18. b) If coupons are given out, then we estimate the average increase in sales is 0.1 units, or $100,000. A 95% confidence interval is: 0.1 ± (2.101)(0.075) = 0.1 ± 0.158 = (−0.058,0.258). The expected effect of coupons on sales is somewhere between a decrease of $58,000 and an increase of $258,000. c) Predicted sales are Y=0.376+0.127(4.7)+0.016(25)+0=1.3729 or $1,372,900. d) For $20,000, the impact on sales of TVAD is (2)(0.127)=0.254, for NEWSAD is (20)(0.016)=0.32 and for coupons it is 0.10. Therefore the most cost-effective medium is NEWSAD. Expected sales increase is $320,000. e) The t-stats for the three media are: for TVAD it is 0.127/0.017=7.47, for NEWSAD it is 0.016/0.003=5.33, and for COUPONS it is 0.100/0.075=1.33. Of these, coupons has the lowest t-stat (in absolute value), with a corresponding p-value of 20% (see the t- table with 18 degrees of freedom and double it). This medium may not have a significant effect on sales. f) The test statistic is 003.0 01.0016.0 − =2.0 and the critical value from the t-table for 18 degrees of freedom is 1.734. Therefore we reject H0. At the 5% level, we conclude that the effect of every $1,000 of NEWSAD on sales is most likely greater than $10,000. g) Note that the R2 is high at 81.5%, and two of the variables have a very significant relationship with sales. However, with an se of 0.1575, any prediction of sales will be accurate to about plus or minus 2(0.1575)=0.315 or $315,000, which limits the utility of the model. 5 thresholds of ±2.575, therefore we accept the null hypothesis. It is still likely that each year of employment at Alphatech increases your salary by $950. The p-value of this test is: 2(0.5-0.4788)=4.24%. e) Plugging in the numbers: SALARY = 24.6 + 0.656(4)- 0.115(0) + 1.90(9) - 1.47(1)=42.854, or $42,854. A 95% confidence interval for the average salary of a person with these attributes would be: 42.854 ± (1.96) 46 823.5 = 42.854 ± 1.683 = (41.171, 44.537), that is, somewhere between $41,171 and $44,537. 3. a) A 99% confidence interval for µ=the true average summer income is given by: 33.1 ± 2.575 45 0.5 = 33.1 ± 1.9. b) Note that the interval from 31.5 to 33.8 is not symmetric around the sample mean 33.1. Therefore, we must solve this problem in two parts: first determine the content in the lower part of the interval (from 31.5 to 33.1) and then the content in the upper part of the interval (from 33.1 to 33.8). The standard error is: 5.0/√45 = 0.75. For the lower part, P(31.5 < X < 33.1) =
   <<− 0 75.0 1.335.31 ZP =P(−2.133 < Z < 0 ) = 0.4834. For the upper part, P(33.1 < X < 33.8) = 
   −<< 75.0 1.338.330 ZP =P(0 < Z < 0.933) = 0.3238. So, the probability content of the interval is 0.4834 + 0.3238 = 0.8072 = 80.72%. 4. a) (A) Multiple R = .984.0969.0 ==− squaredR (B) Degrees of freedom for Regression = K = 3. (C) SSR = 29,657.8−916.7 = 28,741.1. (D) Degrees of freedom for Residual = n−K−1 = 14−3−1=10. b) The test is: H0: β3 = 0 vs. HA: β3 ≠ 0. At the 5% level, we calculate our t-stat, it is 0.0859/0.0379=2.266. So, using the t-table with n−K−1=10 degrees of freedom, the threshold values are ±2.228. Therefore we are (slightly) outside the acceptance region and therefore REJECT the null hypothesis (that β3=0) and conclude that disposable income does have an effect on sales. c) Each unit of disposable income has what effect on sales of Crest? With 95% confidence we are sure it is somewhere between 0.0859 ± 2.228(0.0379) = (0.00146,0.17034) or somewhere between $1.46 million and $170.34 million. Therefore an increase in $3 6 billion in disposable income would increase sales of Crest somewhere between (3)($1.46)=$4.38 million and (3)($170.34)=$511.02 million. This we are 95% sure of. d) This is just a matter of plugging in the right numbers into the regression equation. The equation is: Y = 34.1046 + 3.7459 X1 +30.0463 X2 + 0.0859 X3. Therefore, in this example, we have X1 = 250, X2 = 1 and X3 = 2,000 (remember the units). So, Y = 34.1046 + 3.7459(250) +30.0463(1) + 0.0859(2,000)=1172.4. Or $1.172 billion in sales. e) A 90% prediction interval for the actual sales of part d) is given, in millions of $, by the following (remember we are using the t-table with 10 degrees of freedom): 1,172.4 ± (1.812)(9.574) = 1,172.4 ± 17.3. Therefore, there is a 90% chance that sales were between $1,155.1 and $1,189.7 million. f) Again, this is just a matter of plugging in the right numbers into the equation. In this case, we have X1 = 300, X2 = 300/250 = 1.2 and X3 = 2,000 (remember the units). So, Y = 34.1046 + 3.7459(300) +30.0463(1.2) + 0.0859(2,000)=1365.7. Or $1.366 billion in sales. 5. A 90% confidence interval for the true mean length of life if given by: 5,200 ± (1.860) 9 150 = 5,200 ± 93.00.