Answers to Midterm 2

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PINK AND PURPLE FORMS

1. Answer 1.24% Solution: root(1444) * root( .33 * .67) ----------------------------- * 100 1444 2. 33% plus or missus 3.72% where 3.72% is approximately 3 * 1.24%. 3. The survey is accurate within plus or minus 3.72% for samples of size 1,444 and in this example it gave a 29.28% to 36.72% interval at a confidence level of 99%. What this suggests is that if a sample of size 1,444 were gathered 100 times, the results for each group of respondents would each vary no more than 3.72% in either direction from the parameter in 99 of 100 times. Our hope is that the confidence interval we gave is one of the 99 "good intervals" and not a bad one. 4. The population is All households or all American households The population parameter is the % who will be watching the game. The sample is 1500 households The sample statistic is 6.7% or 100/1500. 5. The answer was 36.195% Z for 22% is 22 - 21 -------------------- 100 * [root(49) * root(.21 * .79)] / 49 and is .172 or .15 (rounding down) Z for 16% is 16 - 21 --------------------------------------- 100 * [root(49) * root(.21 * .79)] / 49 and is -.859 or about -.85 a table look up gives 11.92% + 60.47% and divide this by 2 to get 36.195. 6. The telephone poll is better because it uses a RANDOM PROBABILITY METHOD. THe internet survey involves SELECTION BIAS. Problems are (a) only people with internet access can respond (b) only people who choose to respond will respond, most will ignore these kinds of surveys (c) a respondent can respond as many times as he or she wants, there is no control. 7. The expected value was 0. There are lots of ways to solve this one, basically the box average was 0. 8. The standard deviation of the box = 6.718 the best way to do this was to expand the box...in other words treat it as if there were 10 tickets in there, 5 of them had a +3, 4 of them had a +1.25 and 1 has a -20 so 3, 3, 3, 3, 3, 1.25, 1.25, 1.25, 1.25, -20 and calculate the SD of this list. By the way, you coul calculate the average of this list and it would be the box average. The SE for the sum is root(16) * 6.718 = 26.87 9. The answer here is 18.41% The Z for a score of 25 or more is 25 - 0 ------ 26.87 that gives a .93 or about .90, a table lookup gives 63.19, which is the percentage between +.90 and -.90 so to get the answer, you need to subtract it from 100 and divide by 2.

BLUE AND YELLOW FORMS 1. The answer was 28.2% Z for 34% is 34 - 29 -------------------- 100 * [root(36) * root(.29 * .71)] / 36 and is .66 or .65 (rounding down) Z for 28% is 28 - 29 --------------------------------------- 100 * [root(36) * root(.29 * .71)] / 36 and is -.13 or about -.10 a table look up gives 48.43% + 7.97% and divide this by 2 to get 28.2 2. The population is All households or all American households The population parameter is the % who will be watching the game. The sample is 1200 households The sample statistic is 10% or 120/1200. 3. The telephone poll is better because it uses a RANDOM PROBABILITY METHOD. THe internet survey involves SELECTION BIAS. Problems are (a) only people with internet access can respond (b) only people who choose to respond will respond, most will ignore these kinds of surveys (c) a respondent can respond as many times as he or she wants, there is no control. 4. The expected value was 17.5. There are lots of ways to solve this one, basically the box average was .7. And then, 25 * .7 = 17.5 5. The standard deviation of the box = 5.544 the best way to do this was to expand the box...in other words treat it as if there were 10 tickets in there, 4 of them had a +5, 4 of them had a +1.75 and 2 had a -10 so 5, 5, 5, 5, 1.75, 1.75, 1.75, 1.75, -10 , -10 and calculate the SD of this list. By the way, you coul calculate the average of this list and it would be the box average. The SE for the sum is root(25) * 5.544 = 27.720 6. The answer here is 12.51% The Z for a score of 50 or more is 50 - 17.5 ------ 27.720 that gives a 1.17 or about 1.15, a table lookup gives 74.99, which is the percentage between +1.15 and -1.15 so to get the answer, you need to subtract it from 100 and divide by 2. 7. Answer 1.047% Solution: root(1225) * root( .16 * .84) ----------------------------- * 100 1225 8. 16% plus or minus 2.095% where 2.095% is approximately 2 * 1.047%. 9. The survey is accurate within plus or minus 2.095% for samples of size 1,225 and in this example it gave a 13.91% to 18.095% interval at a confidence level of 95%. What this suggests is that if a sample of size 1,225 were gathered 100 times, the results for each group of respondents would each vary no more than 2.095% in either direction from the parameter in 95 of 100 times. Our hope is that the confidence interval we gave is one of the 95 "good intervals" and not a bad one.

GREEN AND WHITE FORMS 1. The expected value was 36.8 There are lots of ways to solve this one, basically the box average was 2.3 and you multiply this by 16 to get 36.8. 2. The standard deviation of the box = 11.626 the best way to do this was to expand the box...in other words treat it as if there were 10 tickets in there, 6 of them had a +10, 2 of them had a +1.5 and 2 have a -20 so 10, 10, 10, 10, 10, 10, 1.5, 1.5, -20, -20 and calculate the SD of this list. By the way, you could calculate the average of this list and it would be the box average. The SE for the sum is root(16) * 11.626 = 46.50 3. The answer here is 8.85% The Z for a score of 100 or more is 100 - 36.8 ------ 46.50 that gives a 1.36 or about 1.35, a table lookup gives 82.30 which is the percentage between +1.35 and -1.35 so to get the answer, you need to subtract it from 100 and divide by 2. 4. Answer 1.185% Solution: root(1369) * root( .26 * .74) ----------------------------- * 100 1369 5. 26% plus or minus 3.56% where 3.56% is approximately 3 * 1.185%. 6. The survey is accurate within plus or minus 3.56% for samples of size 1,369 and in this example it gave a 22.44% to 29.56% interval at a confidence level of 99%. What this suggests is that if a sample of size 1,369 were gathered 100 times, the results for each group of respondents would each vary no more than 3.56% in either direction from the parameter in 99 of 100 times. Our hope is that the confidence interval we gave is one of the 99 "good intervals" and not a bad one. 7. The answer was 26.56% Z for 35% is 35 - 34 -------------------- 100 * [root(25) * root(.34 * .66)] / 25 and is .11 or .10 (rounding down) Z for 28% is 28 - 34 --------------------------------------- 100 * [root(25) * root(.34 * .66)] / 25 and is -.63 or about -.60 a table look up gives 45.15% + 7.97% and divide this by 2 to get 26.56% 8. The population is All households or all American households The population parameter is the % who will be watching the game. The sample is 1200 households The sample statistic is 7.25% or 87/1200 9. The telephone poll is better because it uses a RANDOM PROBABILITY METHOD. THe internet survey involves SELECTION BIAS. Problems are (a) only people with internet access can respond (b) only people who choose to respond will respond, most will ignore these kinds of surveys (c) a respondent can respond as many times as he or she wants, there is no control.