Question #1: (10 marks) Lecture week 7, Lecture 1

Staph Infections....Staphylococcus aureus is the bacterium that causes staph infections (at least Google says so...) .

Antibiotic resistance in S. aureus was uncommon when penicillin was first introduced in 1943. Indeed, the original Petri dish on which Alexander Fleming of Imperial College London observed the antibacterial activity of the Penicillium fungus was growing a culture of S. aureus. By 1950, 40% of hospital S. aureus isolates were penicillin-resistant; by 1960, this had risen to 80%! Today, S. aureus has become resistant to many commonly used antibiotics. In the UK, only 2% of all S. aureus isolates are sensitive to penicillin, with a similar picture in the rest of the world.

So here's the problem...five different strains of S. aureus was isolated and we wanted to find out if a new antibiotic was at least effective on one of them. So we gathered these five bacteria cultures and placed them in Petrie dishes and watched their survival/extinction rates. Each Petrie dish started with 100 million bacterium of that particular strain. After one week and after the bacteria were kept under ideal growing conditions, the counts were updated. The following table gives the bacteria count, in millions, for the different strains.

Strain A

Strain B

Strain C

Strain D

Strain E

Grand

X

X²

X

X²

X

X²

X

X²

X

X²

Totals

32

1,024

25

625

28

784

32

1,024

31

961

33

1,089

16

256

43

1,849

38

1,444

18

324

20

400

45

2,025

47

2,209

25

625

37

1,369

34

1,156

33

1,089

41

1,681

50

2,500

35

1,225

6

36

19

361

54

2,916

22

484

28

784

39

1,521

39

1,521

25

625

19

361

19

361

49

2,401

T_c

125

196

252

192

217

982

_c

5

7

6

6

7

31

X²

3,705

6,238

10,960

6,702

7,425

35,030

Mean

25.0

28.0

42.0

32.0

31.0

St Dev

12.0

11.2

8.7

10.6

10.8

Can one conclude from the above information and at the 0.05 significance level, that there is a significant difference in responses by the different bacteria stains of S. Aureus to this new antibiotic? Assume the counts in each strain follow a normal distribution and that the variance in the data are the same.. [One would assume there is a difference, but your task as a statistician is to demonstrate this from the acquired data. For example, just by looking at the mean count for Strain A, the antibiotic has an effect whereas for Strain C, it had no effect or minimal effect.]

a)State the Null and Alternative Hypotheses. (1)

b)Construct the probability density curve and state the Decision Rule. (1,1)

c)Generate the ANOVA table to determine the test statistic. (5)

d)What is your decision? (1)

e)What is your interpretation? (1)

Question #2. (7 marks)Lecture Week 7, Lecture 1

So again looking at the data of Question #1, it seems obvious that the bacteria count for Strain A (an average of 25.0 million counts) is much different from that of Strain C (an average of 42.0 million counts). Using the Bonferroni confidence interval approach and at the 0.05 significance level, can one conclude that indeed this is the case; i.e. the effect of the new antibiotic on Strain A is different from that on Strain C.

a)State the Null and Alternative hypotheses (1)

b)State Decision Rule (1)

c)Perform the test (i.e., develop the confidence interval) (3)

d)What is your decision? (1)

e)What is your Interpretation? (1)

Question #3. (22 marks)Lecture Week 5, Lecture 2

a)A cheese company wants to estimate the mean cholesterol content of its soft cheeses. They want their estimate to be within 0.5 milligrams (mg) and they assume the standard deviation of cholesterol in their soft cheeses to be 2.0 mg. If it is assumed that the distribution of cholesterol in their cheese follows a normal distribution, determine the minimal size sample required to construct a 95% confidence interval for the mean concentration of cholesterol in their cheeses. (2)

b)In the palliative care ward, a young doctor conducted a survey to determine the average length of stay of patients. The doctor reported that the 99 percent confidence interval for the mean stay period ranged from 86 to 94 days. S/he was sure that the sample standard deviation was 13 days and that the sample size was at least 30. But s/he could not remember the sample size. From this information, you should be able to help him/her out. (2)

C)From Health Canada, the latest statistics on COPD rates show that for 2013, the annual national rate is 4% and that the provinces with the lowest rates are Manitoba and Saskatchewan at 2.8% and 2.9%, respectively.

Statistics Canada. 2014. Health Trends. Statistics Canada Catalogue No. 82-213-XWE. Ottawa. ReleasedJune 12, 2014.

http://www12.statcan.gc.ca/health-sante/82-213/index.cfm?Lang=ENG.

a)Here in Ontario, our Minister of Health wants an updated estimate for the province. We know that the Ontario estimate will be very close to the national number so using this figure, and using the 95% confidence for our estimate and a reasonable acceptable error of 0.5%, how many people must be surveyed? (4)

b)Now using the results of (a), a survey was conducted and it was found that 206 people in the sample had COPD. From this result, test the claim at 95% confidence that the result from Ontario was different from that of the national average.

i.State the Null and Alternative Hypothesis (1)

ii.Prepare the PDF and state the Decision Rule (1,1)

iii.Compute the test statistic (2)

iv.What is the decision (1)

v.What is your interpretation? (1)

vi.What is the P-value? (2)

c)From the sample information of part (b), determine the 95% confidence interval. Does this support your decision/conclusion from part (b)? Explain. (2,1)

d)Is it possible to state that the current rate for Ontario is even smaller than the posted rates for Manitoba and Saskatchewan in 2013? That is, Ontario had the lowest COPD rate in the country. Explain (1,1)

i.

Question #4. (10 marks) Lecture Week 6, Lecture 1

There have been many studies trying to link the association between different biomarkers and environmental measures of second hand smoke (SHS). Cotinine is the gold standard biomarker for airborne nicotine and in particular for SHS exposure assessment in children with asthma. (Household Smoking Behavior: Effects on Indoor Air Quality and Health of Urban Children with Asthma, Buty, Breysse, et. Al, Matern Child Heath J (2011) 15:460-468)

The bad thing about urine cotinine levels is that it is persistent and stays in the body for extended periods of time. So trying to relate urine cotinine to the number of cigarettes smoked in a home is really quite a hard thing to do because of this latent feature.

A new study was designed to measure the urine cotinine levels of long-distant haulers (truckers) who smoked. The study was trying to get an understanding of SHS in a confined area (the cab of the big rig) and an open area of the home.

Nine truckers were recruited for the study and the results were as follows. The condition being that the truckers had to be driving for a period of at least 8 hours and then their urine was immediately sampled versus the same trucker at home for at least one day before their urine was sampled. The truckers were not to deviate from their smoking habits. In other words, if they smoke six cigarettes every 8 hours in the truck, then they smoked six cigarettes every 8 hours at home.

Cotinine Concentrations (ng/ml)

Driver

Truck Cab

Home

1

295

282

2

286

280

3

289

270

4

425

410

5

345

330

6

405

395

7

365

371

8

424

408

9

283

263

From these data and using the dependent sample analysis approach, test the claim at 95% confidence that the urine cotinine concentrations in the truck cab were higher than those at home.

a)State the Null and Alternative Hypotheses (1)

b)Prepare the PDF and state the Decision Rule (1,1)

c)Determine the test statistic (2)

d)State your decision (1)

e)State the interpretation. (1)

f)Is your decision the same at 99.9% confidence? Explain. (1,2)

Question #5. (27 marks)Lecture Week 5, Lectures 1 and 2

a)A new exercise plan is designed to increase your SBP by 21%. After 300 patients were put through this program, their increased SBP was a whopping 25%. Test the claim at 95% confidence that the plan was a success.

a)What are the Null and Alternative Hypotheses? (1)

b)Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1)

c)Conduct the test(3)

d)State the Decision and Interpretation (1, 1)

e)What is the P_value? (2)

b)Charles Stevens, the owner of Wilmot Orchards, historically had on average of 185 apples per tree for his MacIntosh variety. He applies a new fertilizer to his crop and from a random sample of 36 trees, the average yield is 199 apples per tree with a standard deviation of 48 apples per tree. From these data, test the claim at 95% confidence that the fertilizer actually improved his crop yield.

a)What are the Null and Alternative Hypotheses? (1)

b)Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1)

c)Conduct the test(3)

d)State the Decision and Interpretation (1, 1)

e)What is the P_value? (2)

f)Using this same information, prepare the 95% confidence interval for this new crop. (4)

g)Does the 95% confidence interval contain the older average yield (i.e.,185 apples per tree) before the application of this new fertilizer? (1)

h)If the decision of part i) is to reject the Null Hypothesis, why w/could your confidence interval of part l) include the older average yield of 185 apples per tree? Explain. (2)

Question #6. (14 marks)Lecture Week 6, Lectures 1 and 2

Medical marijuana (well now that I have your attention) is being cultivated by several companies here in Ontario and of course, the companies want their best product put forward. The best product means a product to meet the government standard with respect to potency and yet cost the least to produce (????). After all, they are in it to make money....

In a study of leaf growth, a small number of seedlings were randomly allocated to be grown in either a standard nutrient solution or in a solution containing extra nitrogen. After 30 days of growth, the plants were harvested; the leaf dry weights were determined; and it was noted that the distribution of these data were normal. So, from these data, can one conclude with 99% confidence, that the plants grown in the nitrogen enriched solution had a higher yield than those grown in the regular solution? The data are as follows:

Dry Leaf Weight (grams)

Nutrient Solution

n

Mean

SD

Standard

21

3.85

0.39

Extra Nitrogen

16

4.40

0.71

Since it is obvious that we will be dealing with a t-test statistic,

a)Test the claim at 99% confidence that the variances in the leaf dry weights are not equal.

i.State the Null and Alternative hypotheses (1)

ii.Prepare the PDF and state the Decision Rule (2)

iii.Perform the test (2)

iv.State your Decision (1)

v.State the Interpretation(1)

b)Now test the claim at 99% confidence that the mean weight of the marijuana plants grown in the nitrogen enriched nutrient solution is indeed greater than those grown in the standard nutrient solution.

i.State the Null and Alternative hypotheses (1)

ii.Prepare the PDF and state the Decision Rule (2)

iii.Perform the test (2)

iv.State the Decision (1)

v.State the Interpretation (1)

Question #7. (10 marks)Lecture Week 6, Lecture 2

In 2011, a provincial survey of the long-term care (LTC) sector was undertaken by the Long-Term Care Best Practices Initiative as a follow up to the survey conducted in 2008. The goal of the survey was to evaluate the use of existing resources by Ontario LTC homes in implementing best practices as well as to determine their future needs related to the uptake of best practices in the long-term care sector.

Of the 613 LTC homes in Ontario, 94 participated in the survey resulting in a provincial response rate of only 15%. Those completing the survey were largely in the administrative/management or educator roles. But a large number of front-line bed-side RNs also participated. http://rnao.ca/sites/rnao-ca/files/Highlights_Provincial_Survey_2011.pdf

For the admin/mgmt./educator group, 64 of 80 participating homes reported the LTC Toolkit as their most often utilized RNAO resource. By the way, the RNAO LTC Toolkit is an accessible online repository of resources used by many LTC homes to develop their programs on continence/constipation, fall prevention, pressure ulcers, pain, minimizing restraint use and resident centred care approaches.

For the front-line bed-side RNs, 50 of 72 participating homes also selected the RNAO LTC toolkit as their most often utilized RNAO resource dealing with these best practise issues.

At the 95% confidence level, test the claim that there is a difference in the proportion of the admin/mgmt./educator group and the front-line bed-side RN group using the RNAO LTC toolkit as their most often resource dealing with these best practise issues.

a)What are the Null and Alternative Hypotheses? (1)

b)Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1)

c)Conduct the test(3)

d)State the Decision and Interpretation (1, 1)

e)What is the P_value? (2)

Question #8. (20 marks)Lecture Week 7, Lecture 2 w19

Ok...so "Screen time" is increasing ...especially these last two or three years. We were wondering if this phenomena is quelling the emotions of the younger generation. So we did a small study where we recruited 8 people to participate in four scenarios. The four scenarios were: watching a "Chick Flick"; watching a "Dusty" - an old time Western movie; watching a "Horror" movie; and writing that Midterm Exam #2 in HLSC3800. After the "people were settled in", we measured their systolic blood pressure (SBP). The results were as follows.

Participants

Chick Flick

Dusty

Horror

The Exam

A

125

134

143

128

B

109

112

126

112

C

118

109

124

120

D

132

130

129

132

E

137

137

135

138

F

128

145

148

151

G

115

123

116

128

H

112

118

119

115

Mean

122.0

126.0

130.0

128.0

It seems that writing The Exam has almost the same effect as watching the "Horror" movie...just say'n..J

Anyhow, from a Repeated Measures ANOVA approach, test the claim at 95% confidence that there is a difference in SBP in these four categories. In other words, demonstrate that the participants are aware of their surroundings...again, just say'n. To help you out, I generated this spreadsheet from the collected data.

a)State the Null and Alternative Hypotheses. (1)

b)Construct the probability density curve and state the Decision Rule. (1,1)

c)Generate the Repeated Measures ANOVA table to determine the test statistic. (5)

d)What is your decision? (1)

e)What is your interpretation? (1)

Repeated Measures ANOVA Table

Source of Variation

Sum of Squares (SS)

Degrees of Freedom (df)

Mean Square (MS)

F_test

Between Treatments

Within Treatments

Between Participants

Error

Totals

Furthermore, by gar, use the Modified Bonferroni approach to test the claim at 95% confidence that there is a significant difference between the Chick Flick numbers and the Exam numbers!

f)State the Null and Alternative Hypotheses. (1)

g)Construct the probability density curve and state the Decision Rule. (1,1)

h)Determine the test statistic. (5)

i)What is your Decision? (1)

j)What is your Interpretation? (1)