BIOSTATISTICS: PRACTICE PROBLEM solutions
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These are the solutions to
the set of questions [1-56]
for ENH440
as well as additional questions 56-78 which appear on THIS page below the answers. Just keep scrolling.) For
most questions the
simple numeric answers have been given; you will need to supply the interpretation
for each situation. Some of the more complex
questions will also link to a more detailed solution.
EXACT 'P' RESULTS: In some cases I have given the P value you should have extracted using the tables. In other cases an "exact" P value from a computerized calculation has been given to illustrate the interpretation of such a value. For example, if P is shown as = 0.0347 this still agrees with P < 0.05 but not with P < 0.01. Likewise, P = 0.0017 is clearly < 0.005 but not < 0.001. We WILL be calculating ONE example of an 'exact P' as part of the "Fisher's Exact Test".
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| 1 | OR: 1.94, 1.39, 1.19, 3.26 (CHI-SQ=4.06, P=0.04), 0.63 detailed solution |
| 2 | t = 1.2905, 248 df, P > 0.05 detailed solution |
| 3 | only #4 |
| 4/5 | F(treat) = 2.385, F(age) = 5.474 F(Tr x Age) = 0.987 detailed solution: |
| 6 | OR: 5.00, P=0.215, (FET) |
| 7 | y' = 150.252 - 1.423(x), t= 11.917 |
| 8 | CL: +0.2514, -0.2194 t=0.145, 328.8 |
| 9 | River rafting ... (F.Exact.T) P= 0.0035 Detailed solution |
| 10 | y'= 0.72 + 0.27(x), y=2.88 cm, t=20.0 NEW SCATTERPLOT to illustrate this solution |
| 11 | t=2.429 P<0.05, 76, 127, 178 μgPb/L Detailed solution |
| 12 | |
| 13 | 46 df, P < 0.05 |
| 14 | 7.512 P<0.01, 4.814 P<0.05, 1.951 P>0.05 |
| 15 | t = 3.015, 8 df, P < 0.02 detailed solution |
| 16 | t = 3.953, 38 df, P < 0.001 |
| 17 | t = 1.272, 21df detailed solution |
| 18 | t=2.494, 18 df |
| 19 | t=3.993, 5df |
| 20 | chi-sq = 22.1 detailed solution |
| 21 | t (age) = 1.62, 58 df, P>0.05; t (weight )= 0.74, 58 df, P>0.05; t (bp) = 6.04, 58 df, P<0.001 |
| 22 | t = 0.5606, 11 df, P = 0.59 |
| 23 | t = 0.7845, 4 df, P = 0.527 (this is paired) |
| 24 | t = 0.6978, 11 df, P > 0.05 detailed solution |
| 25 | t = 1.736, 16 df, P >0.05 |
| 26 | Ho: "No difference between Nur and Eng students in terms of mean GPA" P < 0.05 |
| 27 | P > 0.05 |
| 28 | chi-sq 3.525 P = 0.06 OR = 2.85 |
| 29 | chi-sq 6.764 1 df, P < 0.01 RR = 3.3 (protective) |
| 30 | t = 4.202, 18 df, P < 0.001 detailed solution |
| 31 | t = 2.583, 22 df, P < 0.02 detailed solution |
| 32 | (individual calculation) |
| 33 | ms(species) = 73.44, F=4.54 , p>0.05 MS CHEM= 13.083, F = 0.808 p>0.05 |
| 34 | |
| 35 | chi-sq = 7.18, 2df, P<0.05 (Actually P=0.0276) |
| 36 | F = 0.693, 5.403, 6.500 |
| 37 | This is a paired t-test but requires a log transform because the data are exponential t = 1.109, P=0.3306 |
| 38 | MPchi-sq:0.93, P:0.335; HPchi-sq:7.10, P:0.0077(prot); ESFET: P:0.278 |
| 39 | (1) RH r=0.75, y=0.654+0.0892, t=2.971, P<0.05; (2) VEL: r=0.74, y=-0.510+0.0884(x); t=2.879, P<0.05 (3) CORR betw RH&Vel: r=0.22 |
| 40 | please change 6.0 to 60, and 7.0 to 70] Now this is correct: Pb = 233.4 - 31.73(pH) = 233.4 - 31.73(pH) |
| 41 | chi-sq: 4.43, P+0.035, OR 3.18, CL: 0.94 to11.15 |
| 42 | t = 3.315, 7 df, P = 0.0105 |
| 43 | F=10.939, 2,12 df, P=0.002301 detailed solution |
| 44 | (chi-sq goodness of fit)(chi-sq goodness of fit) (You don't need this for the exam) |
| 45 | F(freq):8.44, P<0.01 F(size): 11.25, P<0.01 F(Freq x Size): 4.85, P<0.05 see detailed solution here |
| 46 | b= ―0.1429, CL: ―0.1074 to ―0.1783, t = 34.65, 4df, P <0.001 |
| 47 | d= 2.688, t= 9.622, 7df, P <0.001 |
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48 |
(a) 0.13, (b) 0.81, (c) 0.61, (d) 0.18, (e) >.05, <.002, >.05, >.05 (f) best return in 2 yrs detailed solution |
| 49 | inverse, r=0.58, 10 df, P<0.05 |
| 50 | Fisher's Exact Test P = 0.0204 |
| 51 | đ = 7.4, t = 2.608, P = 0.025 |
| 52 | F= 6.17; P<0.01 Detailed solution |
| 53 | Completed in class. t(unpaired) = 1.077 t(paired) = 8.624 (The correct method is paired) detailed solution |
| 54 | (a) X² 0.47, P=0.49; (b) X² 17.30, (prot) P=0.000032, (c) X² 9.49, P=0.00207 |
| 55 | (a) X² 0.08, P=0.78; (b) X² 0.82, P=0.365, (c) (error:please omit) |
| 56 | X² 6.94, P=0.0084 (or "P<0.01"); OR: 2.37 |
| Page 27 |
[A] F/ratios = 6.37, 22.96, 0.91; detailed solution[B] F/ratios = 15.79, 5.85, 63.16 detailed solution |
ANSWERS TO EXTRA QUESTION SET #57- 81 (Scroll down for these questions)
| 57 | OR: 1.0, 1.17, 0.63, 2.00, 5.44 X?11.1, P=0.0009 (Scroll down for question) |
| 58 | (c) at 250M, Y= 480 ppm, at 500M, Y= 182 ppm, (d) t=9.88 (Scroll down for question) please note the changes here |
| 59 | (a) >0.05 (b) no (d) 26 (Scroll down for question) |
| 60 | t = 9.7, 98 df, ss P<0.001 (Scroll down for question) |
| 61 | RR=4.02 ChiSq: 18.9, 1 df P=0.000014 (P<0.001) Exposed personnel >4x as likely to develop lung ca. reject Ho. (Scroll down for question) |
| 62 | Only expected values are important here. 2 of 9 is 22%, so >20% cells have E values 5 or less. ChiSq not valid |
| 63 | lead(ppm) = 724 - 0.899 M 6df, t = 9.90, P <0.001 (inverse rel. stat sig.) (Scroll down for question) |
| 64 | removed. (essentially the same as #61) |
| 65 | HC% = 47.19 - 0.3695 (ppm LEAD). t = 13.5, 11df, P <.001, r= - 0.97, r2 = 0.94 (Scroll down for question) |
| 70A | SOLUTION: We have here TWO variables. The independent variable is the 'treatment' and is a categorical variable with two levels (a hand barrier cream either antiseptic or not). The dependent variable is presence or absence of E. coli, clearly a categorical variable, with two levels. So the data will be displayed as a 2x2 table. Analysis would be by ChiSquare. (Of course odds ratio would be useful for explaining the strength and direction of the association). |
| 70B | SOLUTION: The independent variable is the same but the dependent variable is not continuous. This is a candidate for either 1-way ANOVA or unpaired t-test, either of which would be applicable. |
| 70C | SOLUTION: This introduces a second independent variable, also categorical, with two levels. If the dependent variable remains as in 70B (continuous), then this is a 2-way ANOVA, and with 240 workers, there will obviously be a number in each group, allowing the "factorial design with interaction". |
| 70D | SOLUTION: Now this takes on a complicated arrangement. If the arrangement in 70B is used, we have a pre-post test (paired t-test) with all 240 people tested before and after using the antiseptic hand cream, thus 240 pairs of data, (239df). But if 70D is used in this way we have TWO independent variables, and the solution is beyond the scope of this course, but might include Raw foods pre-post and Cooked foods pre-post. A t-test in either case would be used. |
| 71 | SOLUTION: The data would be appropriately analysed by means of the paired t-test. Note that the 'pairing' is taking place on the water samples, each of which is being assessed using BOTH tests. Thus the 15 water samples with known lead content produces 30 separate results, or 15 pairs. (14 df) |
| 72 | SOLUTION: t-test for unpaired data. (18 df) |
| 73 | SOLUTION: Chi-square analysis in 2x2 contingency table. True relative risk is appropriate here because you DO have the true incidence data. The two exposure groups were all healthy at the start of the study 30 years ago, and have been followed. Therefore you have the incidence data. RR = Ie/Io or Incidence rate for exposed group over the Incidence rate for non-exposed group. |
| 74 | SOLUTION: Two Variables: Ind.var is categorical with 3 levels. Dep.var is continuous. 1-way ANOVA |
| 75 | SOLUTION: first Ind.var is categorical (3 levels), second Ind.var is also categorical with three levels, Dep.var. is continuous. Two-way ANOVA. Block design if only one obs per group, or Factorial if >1 obs per group |
| 76 | SOLUTION: Chi-square analysis in 3x3 contingency table. But things can get complex. If we do this and just have the total count in each cell, we get a test of the relationship between haz types and training groups. (No pass/fail) So we may stratify - and that is beyond the scope of this course |
| 77 | SOLUTION: Chi-square analysis in 2x3 contingency table. |
| 78 | Click here for detailed solution |
| 79 | F= 23.53, 2, 33 df, P<0.01 detailed solution |
| 80 | F= 3.190, 2,12 df, P>0.05 detailed solution |
| 81 | detailed solution |
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EXTRA QUESTION SET [ #57-80 ]
57. You are investigating risk factors among farm workers for contracting leptospirosis. A group of 30 patients has been identified, as well as a group of 40 non-leptospirosis controls. The following data are the results from enquiries about possible exposures in the previous three months. Data shown are the number stating "yes" to each exposure:
(Q57) |
Have you handled wild animals? |
Do you have a mice infestation? |
Have you visited a zoo? |
Have you handled garden soil? |
Have you repaired sewer pipes or drains? |
cases |
6 |
10 |
4 |
20 |
21 |
controls |
8 |
12 |
6 |
20 |
12 |
58. Data: lead (in ppm) in soil samples (Y) measured at a distance (X) metres from the smoke stack of a lead processing plant. The regression coefficient is -1.1927; the standard error of the regression coeff. is 0.119488; estimated lead concentration at the base of the stack is 778.05 ppm. (a) Plot the data on a scattergram; (b) show the least-squares line; (c) predict the lead in the soil at 250M and 500M distance from the stack; (d) test the null hypothesis of no association; (e) clearly summarize. (Note: you do NOT have to calculate the parameters from the original data.
lead(ppm) y |
40 |
510 |
330 |
160 |
700 |
610 |
220 |
440 |
distance(M) x |
650 |
180 |
405 |
510 |
90 |
190 |
380 |
290 |
59 The results of an investigation into the cadmium content in the blood of children from two areas (A and B) concludes with a statement that the (mean of A) minus (mean of B) was 8.3 µg Cd/100ml blood; 24 df, t=2.01. (a) What is the probability that a mean difference of this amount could have been observed in these two sample groups if there was really no difference between the two areas in terms of children's blood-cadmium? (b) Is this difference statistically significant at the 5% rejection level? (c) Describe clearly what you understand from the results; (d) how many children participated in the study?
60. In correlating haemoglobin levels with estimated exposure to toxic solvents among 100 victims of a chemical spill, the linear correlation coefficient is found to be -0.70 Explain this relationship in detail, including a test of Ho that rho=zero.
61. N=520 men who worked at a Northern Ontario asbestos mill between 1950 and 1970 have been identified through employment records and medical records, and their health/survival status since then is compared with paper mill workers in the same region during the same period. It is found that of the asbestos workers, 25 have developed (or died from) lung cancer. Of 1005 paper mill workers, 12 have been diagnosed (or died from) the disease. (a) Calculate an appropriate risk measurement, and (b) explain what this measurement means.
62. Three cells in a 3x3 contingency table have observed values less than 5, and two of them have expected values less than 5. (a) Is chi-square analysis appropriate here? (b) Why or why not?
63. The effect of distance from an incinerator smoke stack is being investigated. The following data show the lead content in the soil at various distances downwind from the base of the stack. (A) construct a scattergram showing the data and the least-squares line. (B) From your regression model (the equation, not the scattergram) estimate the lead in soil at 650M and 400M. (C) test the null hypothesis that ?zero. (D) summarize fully.
Distance (M) |
100 |
180 |
220 |
340 |
430 |
500 |
610 |
700 |
Lead in soil (ppm) |
680 |
580 |
450 |
420 |
380 |
205 |
200 |
110 |
65. The following data describe the effect of blood lead on haematocrit (packed red blood cells as % of whole blood). Calculate correlation and regression coefficients. Plot the data and the least-squares line on the scattergram. Test the hypotheses (2) that beta=zero and rho=zero
lead µg/dl |
5 |
11 |
13 |
18 |
21 |
27 |
32 |
38 |
41 |
44 |
50 |
58 |
60 |
H.crit% |
45 |
43 |
44 |
39 |
41 |
35 |
37 |
33 |
29 |
32 |
31 |
26 |
24 |
70. What method would you use for these? (70 a-b-c-d-)
70A... A study of effectiveness of new skin antiseptic barrier cream to be used in food manufacturing. Two hundred & forty workers are recruited and randomized into two groups - one group to receive the antiseptic cream, and the other group to receive a similar-looking product without antiseptic action. Each day a hand swab test is taken from the workers and analyzed for E. coli which are reported as "present" or "absent"
70B.... The above study but the outcome (E.coli) is required as a count.
70C...Someone has suggested that those people working with raw foods would be exposed to a much greater bacterial load, so a further division is made between "raw" and "cooked" foods. How would this change the analysis?
70D....As an alternative to the basic study in (70C) above, the investigators contemplate taking all 240 workers and letting them work for a month (with hand swabs every day) before they are asked to use the antiseptic hand cream daily. They are then swabbed daily for another month. What is the statistical analysis you would recommend here?
71... A filter system for removing lead from the water supply is being compared with a conventional filter system. Fifteen water samples with known (but different) concentrations of lead are passed through each of the filters, and the resulting filtrate is examined for lead content. There are 30 samples tested in this way. What method of analysis would you use here?
72....The number of new cases of diarrhoea in infants in a six month period is the outcome measure used in a study of the effectiveness (if any) of a strict double-handwash procedure after changing diapers. Ten centres are selected and invited to participate in the trial, along with another 10 which will undertake normal handwashing practices.
73... In 1975, during the replacement of fuel rods at a nuclear plant, 32 workers were accidentally exposed to radiation for about 4 hours. They have been followed and their health outcomes compared to a group of 48 welders and pipe fitters in a conventional power station. After thirty years, the 8 of the nuclear plant workers and 9 of the conventional plant works have been diagnosed with some form of cancer. What is the appropriate risk measurement? What is the method of analysis to be used?
74 Three methods of training ESL workers on WHMIS procedures are being compared: lecture, power-point, and animated cartoon. The outcome is a knowledge test score out of 40. What method of analysis is to be used here?
75 Same three methods of WHMIS training but this time someone suggests the type of work may play am important part. So the workers are divided into low hazard, medium hazard, and high hazard jobs. What method of analysis is to be used here? ?/fon
76 Same project comparison between three WHMIS training groups x three types of hazard with which the people are working, but this time the dependent var is just ‘pass/fail' the standard test. What method of analysis is to be used here?
?/fo
77 A herbal treatment for contact dermatitis is being evaluated against the conventional treatment of corticosteroids and emollient creams. The outcome after 7 days is recorded as improved? unchanged? or worse? What method of analysis is to be used here?
78 You are investigating the extent to which INDOOR air quality (IAQ) in office buildings without openable windows compares to air quality in similar buildings equipped with windows that open for 5 hours/day. Measurements are taken in three types of EXTERNAL air quality (EAQ). The dependent variable is the count of suspended particles (10 microns or less) per 10 cc. Complete the ANOVA table and summarize.
| GOOD | MODERATE | POOR | |||
| CLO | OPN | CLO | OPN | CLO | OPN |
| N=10 | N=10 | N=10 | N=10 | N=10 | N=10 |
| MEAN=49.6 | MEAN=36.3 | MEAN=49.4 | MEAN=45.6 | MEAN=49.7 | MEAN=54.5 |
| ANOVA TABLE |
| source | SS | df | MS | F | P |
| all groups | 1842.53 | ||||
| EAQ | 810.13 | ||||
| wind O/C | 240.00 | ||||
| interaction | |||||
| residual |
| total |
2570.73 |
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# 79
Three methods of teaching infection control are being compared using three groups of students selected at random. Their scores out of 20 are shown. Is there a significant difference between the means of each group? Explain clearly.
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Method
A: Method B Method C |
5.0 9.0 5.0 8.0 5.0 7.0 6.0 7.0 6.0 6.0 6.0 7.0 12.0 6.0 11.0 7.0 10.0 7.0 10.0 8.0 10.0 9.0 9.0 9.0 3.0 3.0 8.0 7.0 4.0 4.0 6.0 4.0 5.0 5.0 4.0 4.0
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# 80
The pH of three types of canned tomatoes are being compared using the pH readings are shown. Is there a significant difference between the means of each group? Explain clearly.
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Method
1: Method 2 Method 3 |
1.9, 2.4, 2.5, 2.6, 3.0 2.0, 2.1, 3.0, 3.1, 3.3 2.8, 2.9, 3.3, 3.3, 3.8
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NEW
_________________________________________________________________ [a] Enter the missing values in the ANOVA table above [b] Now select the single correct statement that summarizes the analysis A. Old equipm and new equipm perform equally in low pH situations B. New equipm is not affected by pH, but performs consistently better than the old C. In high pH older equipm performs better, but in low pH new equipm is better D. Old equipm and new equipm perform equally in high pH situations E. The performance of the old equipment is not affected by pH
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