Lecture 26 Section 8.4 Hampden-Sydney College Mon, Oct 13, 2008
Outline 1 2 3 4
Exercise 8.12, page 528. Suppose that 60% of all students at a large university access course information using the Internet. (a) Sketch a picture of the distribution for the possible sample proportions you could get based on a simple random sample of 100 students. (b) Use the 68 95 99.7 rule for normal distributions to complete the following statements: (i) There is a 68% chance that the sample proportion is between and. (ii) There is a 95% chance that the sample proportion is between and. (iii) It is almost certain that the sample proportion is between and.
Exercise 8.12, page 528. (c) Would it be likely to observe a sample proportion of 0.50, based on a simple random sample of size 100, if the population proportion were 0.60? Explain. (d) Sketch a picture of the distribution for the possible sample proportions you could get based on a simple random sample of 400 students. (i) How does this picture differ from the one in part (a)? (ii) How will the increased sample size affect the range of values you gave in (i) (iii) of part (b)
Solution (a) For n = 100 students, the sketch is 0.453 0.60 0.747
Solution (b) First, compute µˆp and σˆp. µˆp = p = 0.60. p(1 p) (0.60)(0.40) σˆp = = = 0.0490. n 100 (i) There is a 68% chance that the sample proportion is between 0.60 0.0490 = 0.551 and 0.60 + 0.0490 = 0.649. (ii) There is a 95% chance that the sample proportion is between 0.60 2(0.0490) = 0.502 and 0.60 + 2(0.0490) = 0.698. (iii) It is almost certain that the sample proportion is between 0.60 3(0.0490) = 0.453 and 0.60 + 3(0.0490) = 0.747.
Exercise 8.12, page 528. (c) The question should ask how likely it is to observe a sample proportion at least as low as 0.50. The probability is P(ˆp 0.50), which is normalcdf(-e99,0.50,0.60,0.490) = 0.0206.
Exercise 8.12, page 528. (d) For n = 400 students, the sketch of ˆp is 0.5265 0.60 0.6735
Exercise 8.12, page 528. (d) (i) This distribution is only half as wide (and twice as tall). (ii) The standard deviation of ˆp is only half as much, so the answers are (i) There is a 68% chance that the sample proportion is between 0.60 0.0245 = 0.5755 and 0.60 + 0.0245 = 0.6245. (ii) There is a 95% chance that the sample proportion is between 0.60 2(0.0245) = 0.5510 and 0.60 + 2(0.0245) = 0.6490. (iii) It is almost certain that the sample proportion is between 0.60 3(0.0245) = 0.5265 and 0.60 + 3(0.0245) = 0.6735.
Recall the experiment we did in which we collected 100 samples of size 5 and computed the sample proportions. We graphed our results and saw a good approximation to the normal curve. Then we calculated the mean and standard deviation of our distribution and found that we had good approximations to µˆp = p and σˆp = p(1 p). n Now we will do the same with sample means.
Recall the experiment we did in which we collected 100 samples of size 5 and computed the sample proportions. We graphed our results and saw a good approximation to the normal curve. Then we calculated the mean and standard deviation of our distribution and found that we had good approximations to µˆp = p and σˆp = p(1 p). n Now we will do the same with sample means.
Recall the experiment we did in which we collected 100 samples of size 5 and computed the sample proportions. We graphed our results and saw a good approximation to the normal curve. Then we calculated the mean and standard deviation of our distribution and found that we had good approximations to µˆp = p and σˆp = p(1 p). n Now we will do the same with sample means.
Recall the experiment we did in which we collected 100 samples of size 5 and computed the sample proportions. We graphed our results and saw a good approximation to the normal curve. Then we calculated the mean and standard deviation of our distribution and found that we had good approximations to µˆp = p and σˆp = p(1 p). n Now we will do the same with sample means.
The US Senate There are 100 senators in the US Senate. Their tenures range from 1 year to 49 years. The mean and standard deviation of the population are µ = 13.45 years and σ = 11.18 years.
The US Senate The histogram: 25 20 15 10 5 0 0 10 20 30 40 50
The US Senate The boxplot: 0 10 20 30 40 50
State Years State Years State Years State Years State Years State Years State Years AL 11 DE 7 IA 23 MI 7 NH 5 OK 14 TX 15 AL 10 DE 35 IA 27 MI 29 NH 15 OK 3 TX 6 AK 40 FL 7 KS 11 MN 1 NJ 2 OR 11 UT 31 AK 6 FL 3 KS 12 MN 5 NJ 5 OR 12 UT 15 AZ 13 GA 5 KY 23 MS 1 NM 25 PA 1 VT 1 AZ 21 GA 3 KY 9 MS 30 NM 35 PA 27 VT 33 AR 5 HI 18 LA 11 MO 1 NY 7 RI 1 VA 1 AR 9 HI 45 LA 3 MO 21 NY 9 RI 11 VA 29 CA 16 ID 17 ME 13 MT 1 NC 5 SC 5 WA 7 CA 15 ID 9 ME 11 MT 30 NC 3 SC 3 WA 15 CO 11 IL 11 MD 1 NE 7 ND 16 SD 11 WV 49 CO 3 IL 3 MD 21 NE 11 ND 16 SD 3 WV 23 CT 19 IN 31 MA 46 NV 7 OH 1 TN 1 WI 19 CT 27 IN 9 MA 21 NV 21 OH 9 TN 5 WI 15 WY 1 WY 11
State Years State Years State Years State Years State Years State Years State Years 1 11 15 7 29 23 43 7 57 5 71 14 85 15 2 10 16 35 30 27 44 29 58 15 72 3 86 6 3 40 17 7 31 11 45 1 59 2 73 11 87 31 4 6 18 3 32 12 46 5 60 5 74 12 88 15 5 13 19 5 33 23 47 1 61 25 75 1 89 1 6 21 20 3 34 9 48 30 62 35 76 27 90 33 7 5 21 18 35 11 49 1 63 7 77 1 91 1 8 9 22 45 36 3 50 21 64 9 78 11 92 29 9 16 23 17 37 13 51 1 65 5 79 5 93 7 10 15 24 9 38 11 52 30 66 3 80 3 94 15 11 11 25 11 39 1 53 7 67 16 81 11 95 49 12 3 26 3 40 21 54 11 68 16 82 3 96 23 13 19 27 31 41 46 55 7 69 1 83 1 97 19 14 27 28 9 42 21 56 21 70 9 84 5 98 15 99 1 100 11
Work in pairs. Use the TI-83 to get 10 samples of 5 senators each. (Allow repetitions.) For each sample, find the number of years that each senator has been in the senate. Record the average (out of 5). When you are finished, report the 10 sample means that you found.
Example For example, Sample Tenures {54, 38, 28, 70, 9} {11, 11, 9, 9, 16} 11.2 {46, 84, 6, 72, 49} {5, 5, 21, 3, 1} 7.0 {7, 4, 32, 26, 79} {5, 6, 12, 3, 5} 6.2 {33, 18, 80, 56, 35} {23, 3, 3, 21, 11} 12.6 {85, 54, 59, 25, 27} {15, 11, 2, 11, 31} 14.0 {99, 73, 63, 82, 56} {1, 11, 7, 3, 21} 8.6 {51, 20, 72, 46, 70} {1, 3, 3, 5, 9} 4.2 {70, 1, 93, 87, 95} {9, 11, 7, 31, 49} 21.4 {25, 5, 4, 28, 66} {11, 13, 6, 9, 3} 8.4 {40, 73, 88, 1, 51} {21, 11, 15, 11, 1} 11.8
Example 5 4 3 2 1 0 0 10 20 30 40 50
Study the Central Limit Theorem. Catch up on past homework.