(a) State the null and alternative hypotheses. Will this be a one-tailed or two-tailed test? (b) What is the standardized test statistic?

Question

In golf, the more energy that is transferred from the club to the ball increases the potential distance of a shot. Something called the “coefficient of restitution” (COR) measures this energy transfer. Currently, any golf club with a COR higher than 0.83 is not allowed for official use. Suppose there is suspicion that a certain club maker has exceeded this threshold (which would deem their clubs unusable for tournament play). An experiment is launched to determine if there is evidence to support a claim that the mean COR exceeds 0.83. In this experiment, 20 clubs were selected at random and their CORs measured. This sample produced an average COR of 0.837 with a standard deviation of 0.025. Your job is to perform a hypothesis test for this scenario, following the steps below.

(a) State the null and alternative hypotheses. Will this be a one-tailed or two-tailed test?

(b) What is the standardized test statistic?

(c) Determine if H0 is rejected at level α = 0.05 and give a short, practical interpretation of the result (one or two complete sentences).

Answer 1

Null Hypothesis: μ≤0.83

Alternate Hypothesis: μ>0.83

This is clearly a one-tailed test because there is a specified direction of the mean to be tested.

Since we aren't given the population standard deviation, it is recommended to calculate the t-statistic given the smaller sample size:

t = (x̄-µ)/(s/√n) = (0.837-0.83)/(0.025/√20) ≈ 1.25

Next, we determine the degrees of freedom, which is df = n-1 = 20-1 = 19.

If we refer to a t-table, the t-critical value that corresponds to df=19 and α=0.05 is 1.729 when we find their intersection. Because our t-statistic is less than our t-critical (1.25<1.729), it is outside the rejection region, so we fail to reject the null hypothesis at a significance level of α=0.05.

Therefore, it can be concluded that there is sufficient evidence to support the claim that the mean COR exceeds 0.83.

Hope this helped!