Question

10 marks 4. In 2018 and 2019, Health Canada commissioned two national surveys on the use of cannabis among people aged 20 to 24. They found 175 users out of the 1000 surveyed individuals in 2018 and 230 out of the 1100 sample in 2019 (a) Test the hypothesis that there has been a change in the proportion of cannabis users in the 20-24 age group population between 2018 and 2019. Use the critical value approach and a 0.05 level of significance. (b) Find the p-value for your result in (a) above. (c) Calculate a 95% two-sided confidence interval for the true difference using the data provided (d) Explain how the p-value and the confidence interval are or are not consistent with your result in part (a) above.

Answer 1

Part a)

To Test :-
H0 :- P₂₀₀₈ = P₂₀₀₉
H1 :- P₂₀₀₈ ≠ P₂₀₀₉

p̂1 = 175 / 1000 = 0.175
p̂2 = 230 / 1100 = 0.2091

Test Statistic :-
Z = ( p̂1 - p̂2 ) / √( p̂ * q̂ * (1/n1 + 1/n2) ))
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 175 + 230 ) / ( 1000 + 1100 )
p̂ = 0.1929
q̂ = 1 - p̂ = 0.8071
Z = ( 0.175 - 0.2091) / √( 0.1929 * 0.8071 * (1/1000 + 1/1100) )
Z = -1.9776

Critical value approach
Test Criteria :-
Reject null hypothesis if Z < -Z(α/2)
Critical value Z(α/2) = Z(0.05/2) = 1.96
Z < -Z(α/2) = -1.9776 < -1.96, hence we reject the null hypothesis
Conclusion :- We Reject H0

Part b)
Decision based on P value
P value = 2 * P ( Z < -1.9776 )
P value = 0.048

Looking for the value Z = -1.9776 in standard normal table to find the P value
Reject null hypothesis if P value < α = 0.05
Since P value = 0.048 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

Part c)

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Z(α/2) = Z(0.05 /2) = 1.96
Lower Limit = ( 0.2091 - 0.175 )- Z(0.05/2) * √(((0.175 * 0.825 )/ 1000 ) + ((0.2091 * 0.7909 )/ 1100 ) = 0.0004
upper Limit = ( 0.2091 - 0.175 )+ Z(0.05/2) * √(((0.175 * 0.825 )/ 1000 ) + ((0.2091 * 0.7909 )/ 1100 ) = 0.0004
95% Confidence interval is ( 0.0004 , 0.0677 )
( 0.0004 < ( P2 - P1 ) < 0.0677 )

Since the values in the confidence interval is positive i.e does not contain value 0 in the interval, hence we can reject the null hypothesis.

Part d)

Since all the three methods gives the same result to reject the null hypothesis, hence P value and confidence interval are consistent to our result with critical value method.