Question

Suppose we wanted to know if there is a difference in the proportion of citizens from different provinces who would support the idea of federal laws for mask wearing in public. Upon collecting some sample data we find that 34 out of 50 residents of Ontario would support such legislation, while only 28 out of 50 residents of Alberta support the idea of such laws.

a.) Conduct the appropriate test of hypothesis for these data at the .05 level

b.) What is the p-value associated with these data?

c.) Generate a 90% confidence interval for these data

d.) Why is z always the test statistic used with a proportion, rather than t? (1 mark)

Answer 1

SOLUTION-

LET BE THE POPULATION PROPORTION OF ONTARIO SUPPORTING LEGISLATION AND BE THE POPULATION PROPORTION OF ALBERTA SUPPORTING LEGISLATION. AS PER THE STATED CLAIM, THE HYPOTHESIS IS,

WE USE MINITAB-16 FOR COMPUTATION PURPOSE.

STEPS: STAT> BASIC STATISTICS> TWO PROPORTIONS> ENTER THE SUMMARIZED DATA( FOR SAMPLE 1, ENTER THE EVENT=34 AND TRIAL=40; FOR SAMPLE 2, ENTER THE EVENT=28 AND TRIAL=40)> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 95.0 AND ALTERNATE AS 'NOT EQUAL'> USE THE POOLED P FOR TEST> OK

A.) THE TEST STATISTIC IS Z= 1.61

CRITICAL VALUE= 1.960

AS CRITICAL VALUE EXCEEDS THE TEST STATISTIC, WE FAIL TO REJECT THE NULL HYPOTHESIS.

HENCE, THERE IS NO SIGNIFICANT DIFFERENCE IN PROPORTIONS BETWEEN THE TWO GROUPS.

B.) THE P-VALUE FOR 5% SIGNIFICANCE LEVEL IS 0.180

C.) STEPS: STAT> BASIC STATISTICS> TWO PROPORTIONS> ENTER THE SUMMARIZED DATA( FOR SAMPLE 1, ENTER THE EVENT=34 AND TRIAL=40; FOR SAMPLE 2, ENTER THE EVENT=28 AND TRIAL=40)> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 90.0 AND ALTERNATE AS 'NOT EQUAL'> USE THE POOLED P FOR TEST> OK

THE 90% CONFIDENCE INTERVAL IS (-0.0011, 0.3011)

D.) FOR PROPORTION TESTING, THE SAMPLE SIZE IS GENERALLY HIGHER SO THE DISTRIBUTION APPROACHES NORMAL, AND HENCE, Z-STATISTIC IS PREFERRED.

****IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.