 # 1. Since you are now bio statistical expert you have been asked to ...

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1. Since you are now bio statistical expert you have been asked to give lecture on the subject 17 different classes. You only have time do of these. How many different groups of dasses could you select? Assume that the order notimportant. 2. Suppose 38% of people in: certain population have blood type A. We randomly sample four people from this population (assume these four people are independent). Find the probability that: i) all four have blood type A ii) least one has blood type A (the easiest way to do this sto first find the probability of the complement meaning that none have blood type A) 3. There are 25 patients one wind of small hospital. Of these, 16 are capable of signing an informed consent form, and the other are not. i) It we randomly select one patient, what isthe probability s/he capable of signing? i) If we randomly select two patients (without replacement), what the probability that both are capable of signing? It will be easiest to use the general multiplication rule here. iii) we randomly select four patients (without replacement), what the probability that all four are capable of signing? Here, you1 have to extend the general multiplication rule. 4. screening test for kidney cancer given to 384 people. The results of the test, as well as the definitive diagnosis, are given in the table below. Kidney No Kidney Test Cancer Cancer Total Positive 137 145 Negative 38 201 239 Total 175 209 384 Using this data, calculate (really, estimate) the sensitivity and specificity of the screening test 5. In each situation below, numeric variable is used asa screening tool. For each, state whether relatively large value or a relatively small value would bea "positive" screen. Briefly justify each answer. i) Time to exhaustion on treadmill, using this as screening tool for CVD. i) Body mass index (BMI weight/(height)²), using this as screening tool for diabetes iii) White blood cell count, using this as screening tool for HIV. 6. We want to determine what PSA level should be used to define "positive" screen for prostate cancer. We have the PSA levels of 100 men who have prostate cancer, and another 200 who not. Using this information, we determine the sensitivity and specificity at several potential cutoff values, with the results below. Cutoff value Sensitivity Specificity 2.5+ .940 .460 2,75 - .910 .535 3.0- .900 .620 3.25 + 860 675 3.5 + 840 765 3.75- .800 825 4.0 + .730 .900 4.25 .630 .930 4.5+ 560 940 4.75 - 510 .955 5.0 + .430 .970 i) Sketch an ROC curve to display this information. ii) Which cutoff value do you recommend? Why do you choose this value? 7. screening test has Se 95 and Sp .82. We also believe the prevalence of the disease in some population of interest is .02 (i.e., 2% of the population have the disease). i) Calculate and interpret the PPV for this population. ii) Change the specificity of the test from 82to 92. Recalculate the PPV, and compare to the PPV part (i). What led the change PPV change the number of true positives, change in the number of false positives, or both? iii) Use the original Se of .95 and Sp of .82, but now change the prevalence from 02to .10 Recalculate the PPV, and compare tothe PPV in part (i). What led the change in PPV a change in the number of true positives, change in the number of false positives, or both? 8. Total cholesterol recorded for 30 females and 20 males: Females: 168 172 178 181 183 184 186 188 189 192 192 193 194 194 196 196 197 199 200 200 201 202 204 205 205 208 209 212 215 223 Males: 186 192 197 199 200 202 204 206 207 209 210 220 227 235 239 239 245 247 251 263 i) Using the 30 females only create frequency table with columns for the total cholesterol, f,rif, cf, and rcf. Use classes 160 179, 180 199, 200- 219, and 220 239. For and rdf, round entries to decimal places. ii) For each sex, calculate the median. iii) Descriptive statistics are given below. Use some of this this information to calculate the interquartile range for each sex. Female: Mean 195.533 Std. Dev. 12.409 = 187.50 UQ=204.25 Male: Mean 218,900 Std. Dev.=22.492 LQ=200.50 UQ=239.00 iv) Below is: side-bv-side box plot for this data obtained from SAS. Using this. compare the data sets for the two sexes. In doing so, you should address three distinct things: location, variation, and general shape Justify your observations using statistics from parts (ii) and (iii). 9. State whether or not you think the graphic below is accurate (i.e., accurately depicts the numbers). you sav ves'. briefly explain why. If you say "no', point out at least one thing that makes it inaccurate or misleading Welcometo Microsoft Edge Faster than both Chrome and Firefox benchmark Fastest

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1. Since you are now a bio statistical expert, you have been asked to give a lecture on the subject in 17 different classes. You only have time to do 3 of these. How many different groups of 3 classescould you select? Assume that the order is not important.

The different groups of 3 classes we could select = 17C3 = 17!/(17-3)!*3!
=(17*16*15*14!)/(14!*3!) = 17*16*15/3*2 = 17*8*5 = 680

Hence, we can say that there are 680 different groups of 3 classes we could select.

2. Suppose 38% of people in a certain population have blood type A. We randomly sample four people from this population (assume these four people are independent).
Find the probability that:

i) all four have blood type A

Here we have p = 0.38, n = 4, x = 4
We need to find probability that
P(X=4)
b(x; n, P) = nCx * Px * (1 - P)n - x
=4C4*0.38^4*(1-0.38)^(4-4)
= 0.02085136

So, the probability that all four have blood type A is 0.02085136.

ii) at least one has blood type A (the easiest way to do this is to first find the probability of the complement, meaning that none have blood type A)

Here we have p = 0.38, n = 4, x = 0
We need to find probability that
1-P(X=0)
b(x; n, P) = nCx * Px * (1 - P)n - x
=1-(4C0*0.38^0*(1-0.38)^(4-0))
= 1-0.14776336
= 0.85223664

So, the probability that at least one has blood type A is 0.85223664....

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