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 66 4,356 72 5,184 46 2,116 23 529
50 2,500 44 1,936 50 2,500 53 2,809 35 1,225
64 4,096 42 1,764 40 1,600 45 2,025 50 2,500
34 1,156 54 2,916 42 1,764 55 3,025 63 3,969
22 484 76 5,776 76 5,776 82 6,724 20 400
71 5,041 75 5,625 81 6,561 19 361
30 900 25 625
```

Tc 202 383 355 362 235 1,537

ηc 5 7 6 6 7 31

∑X2 9,260 22,689 22,449 23,260 9,609 87,267

Mean 40.4 54.7 59.2 60.3 33.6

St Dev 16.6 17.0 17.0 16.8 16.9

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 E, the antibiotic has an effect whereas for Strain D, it had little 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. (8 marks) Lecture Week 7, Lecture 1

So again looking at the data of Question #1, it seems obvious that the bacteria count for Strain D (an average of 60.3 million counts) is much different from that of Strain E (an average of 33.6 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 D is different from that on Strain E.

a) State the Null and Alternative hypotheses (1)

b) State Decision Rule (1)

c) Perform the test (i.e., develop the confidence interval) (4)

d) What is your decision? (1)

e) What is your Interpretation? (1)

Question #3. (12 marks) Lecture Week 5, Lecture 2

a) So Mr Ford, Premier of Ontario, would like to know his popularity is at this time… He knows from the last gallop poll in May of 2018 before he introduced the Education Reform items, he had 26% of the popular vote; an all-time low. Using this data, you are to conduct a survey to determine what his popularity is at this time. You must estimate the number of people to be interviewed (sample size) such that the result will be within 3% of the expected proportion 19 out of 20 times. So what is the sample size? (3)

What would be the required sample size if they did not have the information from the last gallop poll? (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 98 percent confidence interval for the mean stay period ranged from 40 to 45 days. S/he was sure that the sample standard deviation was 7.6 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. (3)

C) In 2014, a reproductive clinic in Toronto reported a success rate of 30%. In 2013, they reported 112 live births to 400 women under the age of 40 who had previously been unable to conceive. From this information,

i. Determine the 90% confidence interval for the success rate at this fertility clinic. (2)

ii. Do these data refute or support the clinic’s claim of a 30% success rate? Explain. (1,1)

Question #4. (19 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.

Ten 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

A 325 273

B 288 280

C 295 260

D 424 413

E 375 297

F 455 386

G 365 410

H 424 406

I 289 261

J 381 415

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

(A) A key indicator of the excellence of SickKids researchers is how they compare to the national average when competing for Canadian Institutes of Health Research grant funding. The Canadian Institutes of Health Research (CIHR) is Canada’s largest health research granting agency. In 2010-11, the national success rate was19 per cent – meaning that for every 100 submissions for funding, on average, 19 were successful through the CIHR. During 2010-11, SickKids’ submitted 133 applicants for grants of which, 32 were funded.

http://www.sickkids.ca/annualreport_old/research-and-learning/research-institute/index.html

Test the claim at 98% confidence that Sick Kids had a higher success rate for funding from CIHR than the national success rate – indicating the research excellence of Sick Kids.

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 Pvalue? (2)

(B) Charlie’s weight program is big in downtown Florenceville, New Brunswick. To demonstrate the effectiveness of his program, Charlie recruited 16 volunteers and asked them to follow his diet plan for the next month. Originally, their mean weight was 185 pounds. After one month on the program, their average weight was 179 pounds with a standard deviation of 10 pounds. With such a large standard deviation and a relatively small decrease in weight, the volunteers were still skeptical as to whether the diet plan actually worked. So you, as the statistician, step in and analyze the data. So, from these data, test the claim at 99% confidence that the diet actually worked.

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

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

h) Conduct the test (4)

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

j) What is the Pvalue? (3)

k) Using this same information, prepare the 99% confidence interval for this diet plan. (3)

l) Does the 99% confidence interval contain the before average weight? Discuss the information / decision made in parts i) and k). (1,1)

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 and if one assumes the data follow a normal distribution, can one conclude with 95% confidence that the plants grown in the nitrogen enriched solution had a higher yield than those grown in the regular standard solution? The data are as follows:

` Dry Leaf Weight (grams)`

Nutrient Solution n Mean SD

Standard 21 3.83 0.44

Extra Nitrogen 16 4.26 0.71

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

a) Test the claim at 95% 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 95% 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 split between the administrative/management or educator roles and the front-line bed-side RNs. 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 Pvalue? (2)