Question

Suppose 470 members of the National Defense and Canadian Forces were surveyed and it was found that 72 per cent were in favour of the role of peacekeeping instead of peacemaking. The general public was also surveyed, and 67 per cent were found to be in favour of peacekeeping. This equates to a zobtained of 2.27. What level of confidence can we have in this zobtained?

Please select one of the followings and list all your calculation step by steps.

A.95% for a two-tailed assessment

B. 95% for a one-tailed assessment

C. 85% for a one-tailed assessment

Answer 1

Solution:

Here, we have to use z-test for population proportion.

We are given

Sample size = n=470,

Sample proportion = P = 0.72

Population proportion = p = 0.67

For this z-obtained we need to have 95% confidence or more than 95% for getting more reliable results.

Part A)

H₀: p = 0.67 versus H_a: p ≠ 0.67 (Two tailed test)

Z = (P – p)/sqrt(pq/n)

Where, q = 1 – p = 1 – 0.67 = 0.33

Z = (0.72 – 0.67) / sqrt(0.67*0.33/470)

Z = (0.72 – 0.67) /0.0217

Z = 2.27

Confidence level = C = 95% = 0.95

α = 1 – C = 1 – 0.95 = 0.05

P-value = 0.0212

(by using z-table)

P-value < α = 0.05

So we reject the null hypothesis

Part B.

H₀: p = 0.67 versus H_a: p > 0.67

C = 0.95, so α = 0.05

P-value = 0.0106

P-value < α = 0.05

So we reject the null hypothesis

Part C

H₀: p = 0.67 versus H_a: p > 0.67

C = 0.85, so α = 0.15

P-value = 0.0106

P-value < α = 0.15

So we reject the null hypothesis

***P-values can be obtained from z-table/excel/Ti-83/84 calculator/Software.

***Approximate rounded figures are used during calculations.