1 (4 points) According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%. In later parts of the question we will ask: a. What is the probability that the stock returns are greater than 0%? b. What is the probability that the stock returns are less than 18%? For this part, answer the following question: Which table will we use to find the area under the normal curve? Question 1 options: F-Table Z-Table K-Table T-Table Save Question 2 (7 points) According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%. In later parts of the question we will ask: a. What is the probability that the stock returns are greater than 0%? b. What is the probability that the stock returns are less than 18%? For this part, answer the following question: What is the value of the test statistic (Z, t, or F) for each part? (Round to 2 decimal digits) Question 2 options: -0.87 in part a, -0.06 in part b -0.67 in part a, 0.06 in part b -0.87 in part a, 0.08 in part b 0.67 in part a, 0.08 in part b Save Question 3 (7 points) According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%. In later parts of the question we will ask: a. What is the probability that the stock returns are greater than 0%? b. What is the probability that the stock returns are less than 18%? For this part, answer the following question: What is the area between the mean and our actual score? Question 3 options: 0.408 in part a, 0.039 in part b 0.308 in part a, 0.032 in part b 0.408 in part a, 0.319 in part b 0.308 in part a, 0.039 in part b Save Question 4 (7 points) According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%. a. What is the probability that the stock returns are greater than 0%? b. What is the probability that the stock returns are less than 18%? Question 4 options: P(X > 0) = 80.7%; P(X<18) = 53.2% P(X > 0) = 86.7%; P(X<18) = 53.2% P(X > 0) = 19.2%; P(X<18) =46.8% P(X > 0) = 29.7% ; P(X<18) =53.2% Save Question 5 (5 points) Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. Which table do we use to find the margin of error? Question 5 options: Z-Table T-Table G-Table F-Table Save Question 6 (5 points) Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. What is the Critical value? (Get 4 decimal digits). Use the table here: http://www.statsoft.com/textbook/sttable.html Question 6 options: 2.5706 3.7074 2.4469 2.3646 Save Question 7 (5 points) Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. What is the margin of error? (Round to two decimal digits) Question 7 options: 29.57 37.69 38.80 41.83 Save Question 8 (5 points) Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. Find the lower and upper limits of the interval. Question 8 options: Lower - 15.449; Upper - 67.174 Lower - 20.492; Upper - 40.876 Lower - 18.925; Upper - 50.741 Lower - 5.449; Upper - 80.837 Save Question 9 (5 points) Change the last value to 27 and re-compute the confidence interval. The numbers now are: 21, 28, 33, 34, 25, 26, and 27 Question 9 options: Lower - 3.52; Upper - 41.929 Lower - 23.52; Upper - 31.909 Lower - 22.32; Upper - 35.861 Lower - 19.73; Upper - 51.994 Save Question 10 (5 points) What is an outlier and how does it affect the confidence interval? Question 10 options: An outlier stretches the interval because it increases the standard deviation An outlier compacts the interval because it increases the standard deviation An outlier stretches the interval because it decreases the standard deviation An outlier compacts the interval because it decreases the standard deviation Save Question 11 (6 points) The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that x(bar) = $315.40 and s = $43.20. Using a significance level of .1, what are the boundaries of the confidence interval? Question 11 options: 302.438 - 326.184 300.618 - 330.182 297.485 - 327.820 282.871 - 338.475 Save Question 12 (6 points) The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that x(bar) = $315.40 and s = $43.20. Using the .10 level of significance, is there evidence that the population mean is above $300? Question 12 options: No, because 300 is below the lower limit of the confidence interval Yes, because 300 is below the lower limit of the confidence interval No, because 300 is in the confidence interval Yes, because 300 is in the confidence interval Save Question 13 (7 points) The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that x(bar) = $315.40 and s = $75. (Note the sample standard deviation is different from the first part). Using the .05 level of significance, is there evidence that the population mean is above $300? Question 13 options: No, because 300 is below the lower limit of the confidence interval Yes, because 300 is below the lower limit of the confidence interval No, because 300 is in the confidence interval Yes, because 300 is in the confidence interval Save Question 14 (6 points) Explain the difference between parts A and C. In both parts we are asked whether or not there is evidence that the population mean was over $300. In part A: x(bar) = $315.40; s = $43.2; Sample size = 25 In part C: x(bar) = $315.40; s = $75; Sample size = 25 Question 14 options: The larger mean adds uncertainty in the sample. The sample may have had some outliers. The smaller mean adds uncertainty in the sample. The sample may have had some outliers. The larger standard deviation adds uncertainty in the sample. The sample may have had some outliers. The smaller standard deviation adds uncertainty in the sample. The sample may have had some outliers. Save Question 15 (5 points) The director for Weight Watchers International wants to determine if the changes in their program results in better weight loss. She selected 25 Weight Watcher members at random and compared their weight 6 months later to weight at the start of the program. The results are in this excel file: Weight Watchers.xlsx (The weight in the column labeled \"After\" represents their weights six months later and \"Before\" represents their weight at the start of the six-month period.) The director used .05 as the significance level. Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What are the Null and Alternate Hypotheses? (Keep the Excel file open for the next parts of the question.) Question 15 options: H0: After - Before > 0; HA: After - Before 0 H0: After - Before 0; HA: After - Before > 0 H0: After - Before = 0; HA: After - Before 0 H0: After - Before 0; HA: After - Before < 0 Save Question 16 (5 points) Refer to the Weight Watchers file here: Weight Watchers.xlsx Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is the t-statistic (t-score)? Question 16 options: 7.087 9.321 -7.087 -9.321 Save Question 17 (5 points) Refer to the Weight Watchers file here: Weight Watchers.xlsx Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is the critical t-value? Question 17 options: 3.645 or -3.645 1.711 or -1.711 5.614 or -5.614 7.861 or -7.861 Save Question 18 (5 points) Refer to the Weight Watchers file here: Weight Watchers.xlsx Use Excel to test. For each paired difference, compute After - Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value. What is your conclusion? Question 18 options: Reject the NULL Hypothesis because the actual value is larger than the critical value Do not reject the NULL Hypothesis because the actual value is larger than the critical value Reject the NULL Hypothesis because the actual value is smaller than the critical value Do not reject the NULL Hypothesis because the actual value is smaller than the critical value Person 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Before After 176 192 185 177 196 178 196 181 158 201 191 193 176 212 177 183 210 198 157 213 161 177 210 192 178 164 191 176 176 185 169 196 172 158 193 185 189 175 210 173 180 204 192 152 200 161 166 203 186 170