AP Statistics / Mr. Hansen
2/1/2005

Name: _________________________

Answers to “Stan Stanford” Questions

1.

If X = # who say yes, then E(X) » np = 30(0.045) = 1.35 females.

 

 

2.

Neither, since the condition of independence is not satisfied. (An SRS is sampling without replacement, which does not give independent trials.) However, since the population (300) is at least 10 times larger than the sample (30), we neglect the lack of independence. The other conditions of the binomial setting are satisfied, namely

§         X counts the number of successes

§         n is fixed (n = 30)

§         2 possible outcomes (acceptance or rejection) on each trial

Therefore, we will use B(30, 0.045) as a reasonable approximation of the true distribution.

 

 

3.

sX = Ö(npq) = Ö(30 · 0.045 · 0.955) = 1.135 females

 

 

4.

If Y = # asked in order to obtain first acceptance, then E(Y) » 1/p = 1/0.045 = 22.222 females.

 

 

5.

Neither, since as before, independence is violated. However, the trials are approximately independent since the population is large. (Note that we cannot compare the population to n this time, since n is undefined for the geometric setting.) The other conditions of the geometric setting are satisfied, namely

§         Y counts the number of trials needed in order to obtain the first success

§         n is unspecified

§         2 possible outcomes (acceptance or rejection) on each trial

Therefore, we will use the geometric distribution with p = 0.045 (which we might call G(0.045), though that notation is not standardized) as a reasonable approximation of the true distribution.

 

 

6.

Select pseudorandom integers between 000 and 999 inclusive, where 000 through 044 denote success and everything else denotes failure. Repeated values are acceptable. Since these are independent trials, the simulation does not faithfully capture the conditions of the problem. However, as noted above, we will neglect the lack of independence in the original problem, since the population is large. For each run of random integers, record how many had to be selected in order to obtain one that was 044 or less. (For example, if success occurred on the 12th integer, record “12” on the tally sheet.) Compute statistics after 20 values have been recorded in a data set.

Here are the results I obtained:

730,109,390,744,072,856,065,919,509,480,831,920,002 (success with subject #13)

608,056,940,589,936,620,403,494,708,872,154,315,811,010 (subject #14)

825,118,213,287,497,510,841,062,022 (subject #9)

769,707,050,727,032 (subject #5)

348,445,393,180,801,101,349,581,068,719,332,045,739,780,351,089,814,919,740,924,138,
665,255,520,702,715,047,534,306,859,263,002 (subject #32)

400,718,831,204,001 (subject #5)

210,256,994,190,990,692,189,427,368,122,361,576,225,476,593,010 (subject #16)

303,613,374,036 (subject #4)

165,013 (subject #2)

635,171,133,895,479,462,924,967,506,010 (subject #10)

144,712,381,206,270,911,182,113,409,220,929,264,853,808,791,090,457,500,048,702,303,
469,256,521,193,042 (subject #26)

348,748,906,858,065,350,274,084,133,526,812,747,701,764,717,494,759,102,608,267,043 (subject #21)

130,750,401,694,909,142,071,971,561,419,294,310,422,885,252,836,964,347,686,214,268,
438,766,461,370,540,082,731,558,543,969,878,526,344,638,758,176,065,611,890,319,795,
810,383,846,452,719,635,413,999,448,689,837,989,717,525,737,307,024 (subject #59)

229,555,354,812,652,512,547,667,637,443,565,518,583,126,986,286,356,174,117,683,880,
103,494,445,441,770,703,989,020 (subject #29)

653,770,798,661,499,840,666,595,536,308,682,936,885,281,816,958,888,787,600,626,570,
990,097,523,097,923,731,976,595,223,351,645,418,387,003 (subject #35)

905,506,038 (subject #3)

849,966,536,240,311,128,688,490,983,555,168,631,439,081,951,925,396,888,049,632,955,
710,909,452,948,616,990,205,836,938,762,037 (subject #32)

837,931,209,097,590,992,454,641,936,595,786,491,114,390,387,509,344,673,726,277,619,
765,357,933,255,039 (subject #26)

591,019 (subject #2)

743,181,377,692,988,340,049,088,795,686,778,243,110,412,958,106,727,914,026 (subject #19)

Data set: {13, 14, 9, 5, 32, 5, 16, 4, 2, 10, 26, 21, 59, 29, 35, 3, 32, 26, 2, 19}

Sample mean = 18.1, an estimate for E(Y)

 

 

7.

Sample s.d. = 14.657, an estimate for sY

 

 

8.

The simulation produced an estimate for E(Y) that was low, more than 4 females below the theoretical value of 22.222.

 

 

9.

By formula on p. 475 of main text, sY = Ö(q/p2) = Ö(0.955/0.0452) = 21.716. The simulation produced an estimate for sY that was more than 7 females too low.