Bardzo tanie apteki z dostawą w całej Polsce **kupic cialis** i ogromny wybór pigułek.

## Maths.usyd.edu.au

MATH1015 (Biostatistics) QUIZ 2A - Week 11 (17 May to 21 May)
This is a closed book assessment. However, you are allowed to use information from the course web
page (formulae sheet, statistical tables, etc.) and the R package to answer any question if necessary.

There are 15 questions each with 5 possible answers. Circle the correct answer with pen.

Refer to the following information to answer questions 1-2Suppose the heart rate (measured in beats/min) in healthy individuals is normally distributedwith mean = 75 and standard deviation = 19.

1. The proportion of healthy individuals that have heart rates above 86 is (approx.)
2. 5% of all healthy individuals have heart rate above:
3. Hospital records show that the distribution of birthweight for recent infants is normal with
mean 3.1 Kg and standard deviation 0.6 Kg. A random sample of 16 infants from this hospitalis to be taken. The probability that the mean birthweight of this sample is within 2.85 and3.25 Kg is (close to):
4. Suppose that X1 ∼ N (5, 22) and X2 ∼ N (2, 6.712) (note that 6.712 = 45) Given that X1
and X2 are independent random variables, P (X1 − X2 > 10.5) is:
5. Suppose you roll a fair die 100 times. Let X be number of times (out of the 100) that the die
shows ’6’ on its upper side. The probability P(20 ≤ X ≤ 25), using the normal approximationwith continuity correction is (closest to):
6. Suppose that X ∼ B(500, 0.7). The exact probability, P (325 ≤ X ≤ 350) using R is:
Refer to the following information to answer questions 7-8
It is known that treatment with fosfomycin/trometamol results in bacteriological cure of90% of the patients diagnosed with urinary tract infection. Suppose a new treatment withtrimethoprim/sulfamethoxazole is proposed and in a sample of 100 patients treated withtrimethoprim/sulfamethoxazole 92 were cured. We are interested in testing whether the newtreatment results in higher cure rate. Let ˆ
p be an estimate of the true proportion of cure rate.

8. The P-value for the above test should be calculated as:
9. A random sample of size 16 from a normal population with mean µ and (unknown) variance
x = 12.3 and variance, s2 = 25. A 90% CI for µ is:
10. Using the t-table, P (|t25| > 2.71) is in:
11. A random sample of size 15 has been taken from a normal population to test H0 : µ = 6
against the alternative H1 : µ > 6,. Given that the sample mean, ¯
standard deviation, s = 1.8, the p value for this test is obtained from:
a) P (t14 ≤ 1.08) (b) P (t15 ≥ 1.08) (c) P (t14 ≥ 1.08) (d) P (t14 ≥ 0.96) (e) P (t15 ≥ 0.96).

12. Suppose that we wish to test the hypotheses H0 : µ = 10
normal population with unknown σ. Based on 36 observations from this population, we find
x = 8 and s = 4. This indicates that the exact P-value of this test using R is:
(a) 0.004877 and the data are consistent with H0.

(b) 0.004949 and we have strong evidence against H0.

(c) 0.002474 and we have strong evidence against H0.

(d) 0.995051 and the data are consistent with H0.

(e) 0.004949 and the data are consistent with H0.

13. A researcher wishes to determine if a pill has the undesirable side effect of reducing the blood
pressure of the user. A study involves recording the initial blood pressures of 15 First Yearfemales. After they use the pill regularly for six months, their blood pressures are againrecorded. The mean and variance of the differences of blood pressures, d (=before (x)-after(y)) are given by ¯
d = 0.80 and sd = 1.98. Let µX and µY be the population mean blood
pressures of all potential users before and after the pill and µD = µX − µY . The P-value fortesting H0 : µD = 0 against H1 : µD > 0 is:
(a) 0.05 and we cannot make a decision.

(b) more than 0.05 and the data are consistent with H0.

(c) more than 0.05 and we have evidence against H0.

(d) less than 0.05 and we have evidence against H0.

Refer to the following information to answer question 14-15
Measurements on the size of the mouths (distance from the center of the pituitary to the ptery-omaxillary fissure in mm) of 27 children at age 8 are taken. The file edkwe235x.txt containsthe girls’ measurements and the file edkwe235y.txt contains the boys’ measurements.

Hint: You can read the data into R with the commands x = scan(file = ’edkwe235x.txt’)and y = scan(file = ’edkwe235y.txt’).

Let µ1 be the average mouth size of all girls at age 8 and µ2 be the average mouth size of allboys at age 8. Assume the data were collected from normally distributed populations.

(a) (19.75, 22.61) (b) (21.57, 24.18) (c) (21.41, 24.34) (d) (19.56, 21.57) (e) (19.56, 22.80).

15. The P-value for testing H0 : µ1 = µ2 vs. H1 : µ1 < µ2 is:

Source: http://www.maths.usyd.edu.au/u/UG/JM/MATH1015/r/QUIZ2a-2010.pdf

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Letters in Applied Microbiology ISSN 0266-8254In vitro antiviral activity of Melaleuca alternifolia essentialoilA. Garozzo1, R. Timpanaro1, B. Bisignano1, P.M. Furneri1, G. Bisignano2 and A. Castro11 Department of Microbiological and Gynaecological Sciences, University of Catania, Catania, Italy2 Department of Pharmacobiology, University of Messina, Messina, Italyantiviral activity, essen