Question

Medical tests were conducted to learn about drug-resistant tuberculosis. Of 142 cases tested in British Columbia, 9 were found to be drug resistant. Of 268 cases tested in Ontario, 5 were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two provinces? Use a significance level α=0.01

a) State the hypotheses.

b) Calculate the test statistic.

c) Use Excel to find the p-value.

Hint: Use NORM.DIST.

d) What is your hypothesis test decision?

Hint: There are two possible decisions: reject H0 in favour of Ha or fail to reject H0

e) What is your conclusion in the context of the application?

Hint: The conclusion should relate back to the question.

Answer 1

Given ,

British Columbia :

Sample size = n₁ = 142

Number of drug resistant cases = x₁ = 9

Sample proportion = = = 9/142 = 0.06338

Ontario :

Sample size = n₂ = 268

Number of drug resistant cases = x₂ =5

Sample proportion = = = 5/268 = 0.01866

a)

Hypotheses:

Two tailed test.

b) Test statistic:

Where,

So, test statistic is ,

c) P-value :

P-value for this two tailed test is ,

P-value = 2*P( z > test statistic ) = 2*P( z > 2.3723 )

P( z > 2.3723 ) = 1 - P( z < 2.3723 )

Using Excel function , =NORMSDIST(Z)

P( z < 2.3723 )=NORMSDIST(2.3723) =0.99116

So, P( z > 2.3723 ) = 1 - 0.99116 = 0.00884

P-value = 2*0.00884 = 0.0177

P-value = 0.0177

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level

Given , significance level = α=0.01

It is observed that p-value( 0.0177 ) is greater than α=0.01

So , fail to reject H0

Conclusion :

There is not sufficient evidence to conclude that there is significant difference between the proportions of drug-resistant cases in the two provinces.