A recent article in MacLean’s wrote, “Just over 45 percent said they would buy f
ID: 3355268 • Letter: A
Question
A recent article in MacLean’s wrote, “Just over 45 percent said they would buy food containing marijuana, with 46 percent saying they would purchase pot-laced baked goods like brownies and muffins if they were legal.” (http://www.macleans.ca/society/majority-of-canadians-support-marijuana-legalization-says-survey/)
(a) This statement was based on a Dalhousie University survey of 1087 people across the country. Test at the 0.05 level of significance whether one can conclude that more than 45% of Canadians would say they would buy food containing marijuana. Use the critical value approach (and a 2-sided alternative hypothesis since MacLean’s made the statement after seeing the sample data). Show your manual calculation of the z-statistic and explain how you would find the p-value.
(b) For testing the same hypotheses, calculate manually an appropriate 95% confidence interval for the proportion of Canadians who would buy food containing marijuana. Does this interval allow you to make the same conclusion as in part (a)? Explain briefly.
(c) Suppose you wanted to take a large enough sample size to enable you to conclude that more than 45% of Canadians would buy food containing marijuana. What sample size would be required?
(d) To test whether the proportion of UOttawa students who would buy food containing marijuana exceeded 45%, you found 8 out of a small sample of 10 randomly selected students who answered yes to the question. Perform the test, using the 0.05 level of significance and show how you would calculate the p-value for this test
Explanation / Answer
H0: P = 0.45
H0: P > 0.45
At alpha = 0.05, the critical value is z0.025 = 1.96
The test stststic z = (p - P)/sqrt(P * (1 - P)/n)
= (0.46 - 0.45)/sqrt(0.45 *0.55/1087)
= 0.66
p-value = P(z > 0.66)
= 1 - P(z < 0.66)
= 1 - 0.7454
= 0.2546
As the p-value is greater than the alpha value(0.2546 > 0.05), so the null hypothesis is not rejected.
b) At 95% confidence interval the critical value is z0.025 = +/- 1.96
the confidence interval is
P +/- z0.025 * sqrt(P * (1 - P)/n)
= 0.45 +/- 1.96 * sqrt(0.45 *0.55/1087)
= 0.45 +/- 0.03
= 0.42, 0.48
As the interval cointains the proportion value 0.45, so the null hypothesis is not rejected.
d) H0: P = 0.45
H1: P > 0.45
sample proportion p = 8/10 = 0.8
The test statistic z = (p - P)/sqrt(P * (1 - P)/n)
= (0.8 - 0.45)/sqrt(0.45 * 0.55/10)
= 2.22
P-value = P(Z > 2.22)
= 1 - P(z < 2.22)
= 1 - 0.9868
= 0.0132
AS the p-value is less than the significance level (0.0132 < 0.05), so the null hypothesis is not rejected.
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