Step 1 :Calculate the sample proportion (p̂) which is the number of successes divided by the sample size: \(p̂ = \frac{136}{167} = 0.8144\)
Step 2 :Substitute the values into the formula for the test statistic: \(Z = \frac{(0.8144 - 0.85)}{\sqrt{(0.85(1 - 0.85)) / 167}}\)
Step 3 :Simplify the denominator: \(Z = \frac{-0.0356}{\sqrt{0.1275 / 167}}\)
Step 4 :Further simplify the denominator: \(Z = \frac{-0.0356}{0.0242}\)
Step 5 :Calculate the test statistic: \(Z = -1.471\)
Step 6 :Calculate the p-value, which is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since this is a one-tailed test, find the area to the left of the test statistic on the standard normal distribution. Using a Z-table or a calculator, the p-value is 0.0708
Step 7 :Compare the p-value to the significance level (α = 0.002). Since the p-value is greater than α, we fail to reject the null hypothesis
Step 8 :The final answers are: Test statistic = \(\boxed{-1.471}\), p-value = \(\boxed{0.0708}\), and the decision is to \(\boxed{\text{fail to reject the null hypothesis}}\)