C) From Health Canada, the latest statistics on COPD rates show that for 2013, t

Question

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. Released June 12, 2014 /www12.statcan ealth-san 13/index.c 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) 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 b) i. State the Null and Alternative Hypothesis (1) ii. Prepare the PDF and state the Decision Rule (1,1) ili. iv. v. vi. What is the P-value? (2) Compute the test statistic (2) What is the decision (1) What is your interpretation? (1) From the sample information of part (b), determine the 95% confidence interval. Does this support your decision/conclusion from part (b)? Explain. (2,1) 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) c) d)

Explanation / Answer

a)

Hence sample size should be 38415.

b)

i) Below are the null and alternate hypothesis
H0: p = 0.04
H1: p not equals to 0.04

ii) This is two tailed test and we reject the null hypothesis if test statistics falls in the rejection region. For 95% CI, rejection regions would be beyond -1.96 and 1.96

iii) pcap = 206/38415 = 0.0054
Test statistics
t = (0.0054 - 0.04)/sqrt(0.04*0.96/206) = -2.534

iv) Reject null hypothesis

v) There is sufficient evidence to conclude that proportion is different from national proportion.

vi) p-value = 0.0113

c) z-value = 1.96
SE = sqrt(0.0054*(1-0.0054)/38415) = 0.00037
ME = z*SE = 1.96*0.00037 = 0.0007

CI = (0.0054 - 0.0007, 0.0054 + 0.0007) = (0.0047, 0.0061)

As value 0.04 does not lie within CI, reject null hypothesis.

d) Ontario does not have smaller rate than others because we find the confidence interval biased towards upper tail.

Margin of Error, ME 0.005 z-value 1.959963985 p 0.5 n = (z/ME)^2*p*(1-p) 38414.58821

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