Suggested problem set #4 Chapter 7: 4, 9, 10, 11, 17, 20 Chapter 7. 4. For the following scenarios, say whether the binomial distribution would describe the probability distribution of possible outcomes. If not, say why not. a. The number of red cards out of the first five cards drawn from the top of a regular deck. No, because the probability of getting a red card would change as the cards are removed. The binomial assumes that the probability of success is equal and independent for each trial. b. The number of red balls out of ten drawn one by one from a vat of 50 red and blue balls, if the balls are replaced after each draw. Yes c. The number of red balls out of ten drawn one by one from a vat of 50 red and blue balls, if the balls are not replaced after each draw. No, for the same reason as in question a. d. The number of red flowers in a square meter plot in a field that has been randomly strewn with an indefinitely large amount of seed. No, because the number of trials is not specified. e. The number of red eyed flies among 50 Drosophila individuals drawn at random from a large population. yes f. The number of red eyed flies in five Drosophila families, each of ten individuals, with the families chosen at random from a large population. no, because the individuals are not chosen at random. 9. a. pˆ =

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1856 =0.323 5743

b. No, because this is not a sample – it describes the entire population. Any error in this estimate (and there is likely some) is due to measurement error, not sampling variation. 10. Ho: The proportion of species moving north is p=0.5. Ha. The proportion of species moving north is not p=0.5. Use a binomial test. P = 2 * (Pr[x = 22] + Pr[x = 23] + Pr[x = 24]) P = 0.000002 We reject the null hypothesis, and conclude that most species are moving north.

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11. a. The first, because the confidence interval is smaller. b. The first, for the same reason c. No, because their confidence intervals overlap, and the confidence interval for the second study includes the mean from the first study. 17. Carry out a binomial hypothesis test. Ho: The probability of choosing an OM female is p=0.5. Ha: The probability of choosing an OM female is different from 0.5. Test statistic: 19 out of 24 choose OM P= 2 * (Pr[x = 19] + Pr[x = 20] + Pr[x = 21] + Pr[x = 22] + Pr[x = 23] + Pr[x = 24]) P = 0.0015

Reject the null hypothesis. There is evidence that male choice is affected by fetal positioning.

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b. Worse, because then the two females in each trial would not be independent. 20. a. 30% b. Binomial c. The same as the standard deviation of a binomial distribution: n * p(1" p) = 15 * 0.3(0.7) = 1.77 n=15 here because it’s the number of cells picked per researcher, and not the number of researchers, that determines the s.d. p(1" p) 0.3(0.7) d. = = 0.118 ! n 15 e. 95% !

Chapter 8: 3, 6, 8, 14 3. a. Six categories – 1, 2, 3, 4, 5, 6. Five degrees of freedom (binomial dist.) b. Data are the number of heads in 10 tries. 11 categories – 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 heads. 10 degrees of freedom (binomial dist.). c. Same number of categories. Here you’d have to estimate the parameter p, so there’s one less degree of freedom – 9. d. 5 Categories: 0, 1, 2, 3, 4. 3 degrees of freedom (5-1-1). 6. a. Ho: Birth rates on weekdays and weekends are equal. Ha: Birth rates differ between weekdays and weekends. b. Binomial: difficult to calculate but exact, assumes random sample Chi-squared goodness of fit test – easier to calculate, requires large sample size to be accurate. c. Ho: Birth rates on weekdays and weekends are equal.

Ha: Birth rates differ between weekdays and weekends. Use chi-squared goodness of fit test, with the proportional model. Observed 716 216 932

Weekdays Weekends Total

Expected 665.7 266.3 932

Expected for weekdays: 5/7 * 932 = 665.7; weekends: 2/7 * 932 = 266.3 Test statistic: (O # E) 2 (716 # 665.7) 2 (216 # 266.3) 2 =3.80+9.50=13.3 "2 = $ = + E 665.7 266.3 df = 2-1 = 1 !

Critical value at alpha = 0.05: 3.84 The test statistic is larger than the critical value, so we reject the null hypothesis. 8. Use a chi-squared goodness of fit test for a Poisson distribution. Ho: Number of parasites follow a Poisson distribution Ha: Number of parasites follow some other distribution. Calculate the mean: (0*103+1*72+2*44+3*14+4*3+5*1+6*1)/238=225/238=0.945 Use this in the Poisson formula to calculate the expected values: Number of parasites 0 1 2 3 4 5 6

Frequency 103 72 44 14 3 1 1

Expected 92.51 87.42 41.30 13.01 3.07 0.58 0.09

This violates the assumptions of chi-squared, so we lump the last three categories: Number of parasites 0 1 2 3 >4

Frequency 103 72 44 14 5

Expected 92.51 87.42 41.30 13.01 3.75

Now calculate the chi-squared statistic: 4.58 df = 5 – 1 – 1 = 3 Critical value with alpha = 0.05, 3 df: 7.81 Fail to reject the null hypothesis – there is no evidence that fish differ in their ability to attract nematodes. 14.

"15% a. Pr[x = 8] = $ '(0.5) 8 (0.5) 7 #8& b. Use a goodness-of-fit test. !

Wins

Observed 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 7 25 56 112 190 290 256 549 339 132 73 56 22 22 3 2132

Binomial 0.00 0.00 0.00 0.01 0.04 0.09 0.15 0.20 0.20 0.15 0.09 0.04 0.01 0.00 0.00 0.00

Expected 0.07 0.98 6.83 29.60 88.81 195.39 325.64 418.68 418.68 325.64 195.39 88.81 29.60 6.83 0.98 0.07

We have to lump the first two and last two categories. Wins 0-1

Observed 2 3 4 5

7 25 56 112 190

Expected 1.04 6.83 29.60 88.81 195.39

Chi-squared 34.11 48.32 23.54 6.05 0.15

6 7 8 9 10 11 12 13 14-15

290 256 549 339 132 73 56 22 25 2132

325.64 418.68 418.68 325.64 195.39 88.81 29.60 6.83 1.04

3.90 63.21 40.56 0.55 20.56 2.82 23.54 33.68 552.00 852.98

The critical value for a chi-squared with 13 df is 22.36. We reject the null; these data do not fit a binomial distribution.

c. You can see from the above plot that the observed number of wrestlers with exactly 8 wins is much higher than expected. d. There is an alternative – that some wrestlers are just better than others. This would mean that for each match, p is not equal to 0.5. The plot provides some support for this in that there are way more contestants with 0-1 wins (very bad) and 14-15 wins (very good) than expected. Chapter 9: 5, 8 5. (84+72)/5865 pairs were of different species.

so:

156 = 0.026 5895 pˆ (1" pˆ ) 0.026(1" 0.026) SE pˆ = = = 0.0021 n "1 5895 "1 pˆ =

Do a chi-squared contingency test.

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Observed: Collared Pied Column sums

Female Collared 5567 72

Pied

5639

84 172

Row Sums 5651 244

256

5895

245.4 10.6

Row Sums 5651 244

256

5895

Expected: Collared Pied Column sums

Test statistic:

Female Collared 5405.6 233.4

Pied

5639

(O " E) 2 # E = 5895 .

Chi-squared distribution with (r-1)(c-1) = 1 d.f. Critical value: 3.84. ! Our test statistic is greater than the critical value, so we reject the null hypothesis. The birds are pairing nonrandomly. 8. a.

Gave birth Didn't give birth

Expected Number of matings 1 82.37 4.63 87

2 88.05

3 57.75

4 16.10

5 4.73

249

4.95 93

3.25 61

0.90 17

0.27 5

14 263

b. No – there are cells with expected values less than one, and more than 20% with expected less than 5. Categories need to be pooled to meet the assumptions. c. Observed Number of matings 1-2 3-5

Gave birth Didn't give birth

83

249

14 180

0 83

14 263

78.58

249

4.42 83

14 263

Expected Number of matings 1-2 3-5 170.42

Gave birth Didn't give birth

Test statistic:

166

9.58 180

#

(O " E) 2 = 6.82 E

Chi-squared distribution with (r-1)(c-1) = 1 d.f. Critical value: 3.84. ! Our test statistic is greater than the critical value, so we reject the null hypothesis. d. No – both could be correlated with a third variable, like female condition. Females who are in better shape – plenty of food, not sick – might get more mates and also have larger litters.