HWK092019 Springanswer key PDF

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Solutions to practice questions for the course...

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STATISTICS 111

HOMEWORK 9 ANSWER KEY

1. A medical researcher interested in the genetic basis of chronic bronchitis collects data from 1000 single-child families where at least one parent suffers from this illness. She finds that 375 of the children also suffer from the illness. She also collected data from 1000 single-child families where neither parent suffers from this illness. She finds that 345 of these children suffer from the illness. Let θ1 be the probability that a child in single child family suffers from chronic bronchitis for the case where at least one parent suffers from chronic bronchitis, and θ2 be the probability that a child in single child family suffers from chronic bronchitis for the case where neither parent suffers from chronic bronchitis. From the above data, find an unbiased estimate of the difference θ1 - θ2 and calculate a conservative approximate 95% confidence interval for θ1 - θ2 . [3 + 5 = 8 points] Estimate is 375/1,000 – 345/1,000 = 0.030.

[3 points]

The (conservative) formula that should be used leads to 0.030 - √{1/1,000 + 1/1,000} = 0.030 – 0.045 = -0.015 to 0.030 + √{1/1,000 + 1/1,000} = 0.030 + 0.045 = 0.075 You also get full points if you use a correct “non-conservative” formula.

[5 points]

2. Suppose the data in question 1 are changed, so that the researcher collects data from 10,000 single-child families where at least one parent suffers from chronic bronchitis and finds that 3700 of the children also suffer from this illness. The researcher also collects data from 10,000 singlechild families where neither parent suffers from chronic bronchitis and finds 3350 of the children suffer from this illness. From these new data, find an unbiased estimate of the difference θ1 - θ2 and calculate a conservative approximate 95% confidence interval for θ1 - θ2 . [3 + 5 = 8 points] Estimate is 3700/10,000 – 3350/10,000 = 0.035. The (conservative) formula that should be used leads to 0.035 - √{1/10,000 + 1/10,000} = 0.035 – 0.014 = 0.021 to 0.035 + √{1/10,000 + 1/10,000} = 0.035 + 0.014 = 0.049

[3 points]

You will be awarded full points if you used a correct “non-conservative” formula. [5 points] 3. Comment on the similarities and differences between your answers to questions 1 and 2, basing your comment around the formula for the mean, and also the formula for the variance, of the difference between two proportions. [7 points] The estimates of θ1 - θ2 are close to each other, and this makes sense because they are both unbiased estimates of θ1 - θ2 . [2 points] However in Question 1 the confidence interval is much wider than that in Question 2, and this arises because of the larger sample size in Question 2. [5 points]

4. We are interested in investigating any potential difference between the mean blood sugar level of diabetics (μ1) and that of non-diabetics (μ2). To do this we took a sample of ten diabetics and found the following blood sugar levels: 127, 144, 140, 136, 119, 138, 145, 122, 132, 129. We also took a sample of eight non-diabetics and found the following blood sugar levels: 125, 128, 133, 141, 109, 125, 126, 122. Find an unbiased estimate of the difference μ1- μ2. Find two numbers between which we are about 95% certain that μ1- μ2 lies.

[4 points] [10 points]

NOTE: Various calculations for the diabetics were done in Homework 8, Question 3. Specifically, the estimate of the mean blood sugar level among diabetics is x = 133.2 and the estimate of the variance of blood sugar levels among diabetics is s2 = 79.7333. You may use these in answering this question. Let x 1 be the average of the diabetic readings and x 2 be the average of the non-diabetic readings. Then we estimate μ1- μ2 by x 1 - x 2 = 133.2 - 126.125 = 7.075.

[4 points]

Find two numbers between which we are about 95% certain that μ1- μ2 lies. The formula that you should use is x 1 - x 2 plus and minus 2 √{s21/n+ s22/m}. Now n = 10 and m = 8. s21 is 79.7333 (see above) and s22 to be 83.5536. This leads to the limits 7.075 8.583 and 7.075 + 8.583, that is -1.508 and 15.658. The important point is to use the formula x 1 - x 2 plus and minus 2 √{s21/n+ s22/m}. Only a few points will be taken off if you used this formula correctly but makes a numerical mistake. [10 points] 5. We wish to estimate the way in which weight (Y) (in kg) of an infant girl between the ages of 12 and 36 months depends on their age (x) (in months). More specifically, we assume that the mean of Y is of the form α + βx, where α and β are two unknown parameters.

A sample of 10 infant girls gave the following data:

Infant # Age Weight

1 33 12.9

2 32 13.8

3 14 8.2

4 20 12.2

5 15 8.5

6 16 12.9

7 30 13.7

8 17 11.2

9 21 11.9

10 23 10.4

Plot these data accurately on a graph. Estimate the mean weight of an infant of age x via the formula “estimated mean weight of an infant of age x is a + bx”. (That is, calculate a and b, where a and b are, respectively, the estimates of α and β.) Plot your line corresponding to the equation y = a + bx on your data graph. Does it “skewer” through your data points? [13 points] [Hint: x = 22.1, y = 11.57, sxx = 464.9, sxy = 93.53, syy = 35.841] You get full points if you used any method for graphing, including by hand, by Excel, etc. [1 point]

Estimate the mean weight of an infant of age x via the formula “estimated mean weight of an infant of age x is a + bx”. (That is, calculate a and b.) Using the formulas given in class, b = 93.53/464.9 = 0.2011831 = 0.201 (Rounding to three decimal place accuracy.)

[2 points]

a = 11.57 – (0.2011831)(22.1) =7.12385 = 7.12. (Rounding to three decimal place accuracy.) [2 points]

Plot your line on your data graph. You can use any method for graphing, including by hand.)

[1 point]

Does it “skewer” through your data points? Yes.

[1 point]

Find two limits between which you are approximately 95% certain that β lies, using the formula given in class. Here you should use formula (102) in the notes. First, from (101), s2r = {35.841 – (93.53)2/464.9}/8 = 2.12804. Thus sr = 1.45878. Then from (102) the lower limit is 0.201 – 2(1.45878)/√464.9 = 0.201 – 0.135

= 0.066

[3 points]

and the upper limit is 0.201 + 0.135 = 0.336.

Total: 50 points

[3 points]...