Philip Beal Statistics set 3 Homework AMBA 600 - Section 9046 Dr. Susan Malone Sep 8, 2015 Problem Set: Question 1 1. The University of Maryland University College is concerned that out-of-state students may be receiving lower grades than Maryland students. Two independent random samples have been selected: 126 observations from population 1 (out-of-state students) and 203 from population 2 (Maryland students). The sample means obtained are X1 (bar) = 99 and X2 (bar) = 101. It is known from previous studies that the population variances are 10.9 and 11.4 respectively. Using a level of significance of .01, is there evidence that the out of state students may be receiving lower grades? Fully explain your answer. Reminder: Please use the six steps for hypothesis testing as outlined in your Lind textbook. Yes, that means I want you to label and discuss each step (Step 1, Step 2, etc.). Step 1: State the Null Hypothesis (H0) and the Alternate Hypothesis (H1) Step 2: Select the level of significance Step 3: Select the test statistic Step 4: Formulate the decision rule Step 5: Make a decision Step 6: Interpret the result Hypothesis: Step 1:Null Hypothesis, Ho: there is no significant difference in the grades of out of state students and finance students. Alternative Hypothesis, H1: the out of state students may be receiving lower grades than finance students. Level of significance: Step 2: Alpha = 0.01 Test statistic: Step 3: T test for comparing means, n1 x1 bar s1^2 126 99 10.9 n2 x2 bar s2^2 203 101 11.4 S^2(pooled) = =((125*10.9)+( 202*11.4))/SQR T(126+203-2) 202.6916 =(99101)/SQRT(202 .6916*((1/126)+ (1/203))) t= Decision Rule: Step 4: If -t > -t(0.05, n1+n2-2), accept Ho at % level of significance, otherwise we may reject it. Decision: Step 5: Since, -t > -t(0.05, n1+n2-2), I accept Ho at 5% level of significance. Interpretation: Step 6: There is no significant difference in the grades of out of state students and finance students Simple Regression 2. A CEO of a large pharmaceutical company would like to determine if the company should be placing more money in the budget next year for television advertising of a new drug for controlling diabetes. He wonders whether there is a strong relationship between the amount of money spent on television advertising for this new drug called DIB and the number of orders received. The manufacturing process of this drug is very difficult and requires stability, so the CEO would prefer to generate a stable number of orders. The cost of advertising is always an important consideration in the phase I rollout of a new drug. Data that have been collected over the past 20 months indicate the amount of money spent of television advertising and the number of orders received. Month Advertising Cost Number of Orders 78,420 62,620 55,580 93,680 71,180 63,140 85,370 76,880 69,990 77,230 59,380 62,750 73,270 86,190 61,530 79,540 63,350 84,530 79,760 74,640 Advertsing Cost 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A. Set up a scatter diagram and calculate the associated correlation coefficient. Discuss how strong you think the relationship is between the amount of money spent on television advertising and the number of orders received. Please use the Correlation procedures within Excel under Tools > Data Analysis. The scatterplot can more easily be generated using the \"Chart\" procedure. 2,856,000 1,810,000 1,299,000 1,730,000 2,367,000 2,611,000 4,778,000 1,935,000 2,155,000 3,434,000 1,598,000 1,867,000 2,899,000 3,545,000 1,534,000 2,891,000 1,625,000 3,778,000 2,979,000 3,814,000 Scatter Plot 100000 80000 60000 40000 20000 0 1000000 1500000 2000000 2500000 3000000 3500000 Number of orders 4000000 4500000 5000000 B. Assuming there is a statistically significant relationship, use the least squares method to find the regression equation to predict the advertising costs based on the number of orders received. Please use the regression procedure within Excel under Tools > Data Analysis to construct this equation. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.1255413177 0.0157606224 -0.038919343 10657.315463248 20 ANOVA df Regression Residual Total Intercept X Variable 1 1 18 19 SS 32737143.1026845 2044410711.89732 2077147855 Coefficients 70055.02309703 0.0007836046 Standard Error 5897.9449873489 0.0014595683 MS 32737143.1026845 113578372.883184 F 0.2882339505 Significance F 0.5979286896 t Stat P-value Lower 95% Upper 95% 11.8778698763 5.94889759E-010 57663.900480979 82446.1457131 0.5368742408 0.5979286896 -0.0022828346 0.0038500438 RESIDUAL OUTPUT Observation Predicted Y Residuals 1 72292.997865351 6127.0021346494 2 71473.347442499 -8853.3474424991 3 71072.925486403 -15492.9254864025 4 71410.659073639 22269.3409263614 5 71909.815210691 -729.8152106906 6 72101.014735715 -8961.0147357152 7 73799.085927225 11570.914072775 8 71571.298018844 5308.7019811563 9 71743.69103321 -1753.6910332102 10 72918.314344735 4311.6856552656 11 74441.641708046 -15061.6417080455 12 72379.977977145 -9629.9779771446 13 72326.692863613 943.3071363869 14 73381.424669692 12808.5753303083 15 73921.328246503 -12391.328246503 16 73074.251662275 6465.748337725 17 74462.799032536 -11112.7990325359 18 74582.690537982 9947.3094620183 19 74755.867156959 5004.1328430411 20 75410.177006941 -770.1770069407 The equation for regression/monthly advertising cost = y-intercept+Slope number of orders C. Interpret the meaning of the slope, b1, in the regression equation. Slope = 0.00078360461075669, With a unit increase in the number of orders, advertising cost will increase by 0.00947739931640059. D. Predict the monthly advertising cost when the number of orders is 2,300,000. (Hint: Be very careful with assigning the dependent variable for this problem.) The monthly advertising cost when the number of orders is 2,300,000 is 70055.0230970295+0.00078360461075669*2300000=$71,857.303 E. Compute the coefficient of determination, r2, and interpret its meaning. The Coefficient of determination is .0158. This means that 1.58% of the advertising variation can be explained by the regression tool F. Compute the standard error of estimate and interpret its meaning. The standard error is 10657.315463248.This is the estimate of error variance G. Do you think that the company should use these results from the regression to base any corporate decisions on? Fully explain your answer. I think the company should not base any corporate decisions on the results from the regression. F =0.288233950457758 with p value =0.597928689550798. the P value is gre eater than .01 level of significance which means the hypothesis can not be rejected 3. Dr. Michaella Evans, a statistics professor at the University of Maryland University College, drives from her home to the school every weekday. She has three options to drive there: She can take the Beltway, or she can take a main highway with some traffic lights, or she can take a back road which has no traffic lights but is a longer distance. Since she is data-oriented, she is interested to know if there is a difference in the time it takes to drive each route. As an experiment, she randomly selected the route on 21 different days and wrote down the time it took her for the roundtrip, getting to work in the morning and back home in the evening. At the .01 significance level, can she conclude that there is a difference between the driving times using the different routes? Null Hypothesis, Ho: there is no significance difference between mean time taken by beltway, main highway and back road. Alternative Hypothesis, Ho: atleast one of the mean time taken by beltway, main highway and back road is significantly different. Anova: Single Factor SUMMARY Groups Count Beltway Main Highway Back road Sum 8 6 7 Average Variance 710 88.75 37.3571428571 496 82.66667 47.8666666667 582 83.14286 107.80952381 ANOVA Source of Variation SS df MS Between Groups Within Groups 168.880952 1147.69048 2 84.44048 18 63.76058 Total 1316.57143 P-value F crit 20 p-value = F 1.3243366595 0.290685 3.554557 0.29068536 since, p-value is greater than alpha (0.05), I fail to reject Ho at 5% level of significane and conclude that there is no significance difference between mean time taken by beltway, m and back road. Time (in minutes) it took to get to work and back using: Beltway 87 93 91 88 99 83 90 79 Main HighwayBack road 80 88 82 80 90 81 92 96 77 95 75 73 69 eltway, main highway