Question

Each of 51 golfers hit three golf balls of brand \(X\) and three golf balls of brand \(\mathrm{Y}\) in a random order. Let \(X_{i}\) and \(Y_{i}\) equal the averages of the distances traveled by the brand \(\mathrm{X}\) and brand \(\mathrm{Y}\) golf balls hit by the \(i\) th golfer, \(i=1,2, \ldots, 51\). Let \(W_{i}=X_{i}-Y_{i}, i=1,2, \ldots, 51 .\) To test \(H_{0}: \mu_{W}=0\) against \(H_{1}: \mu_{W}>0\), where \(\mu_{W}\) is the mean of the differences. If \(\bar{w}=2.07\) and \(s_{W}^{2}=84.63\), would \(H_{0}\) be accepted or rejected at an \(\alpha=0.05\) significance level? What is the \(p\) -value of this test?

Short answer

In this task, the null hypothesis will be rejected or accepted at the 0.05 significance level evaluated as per comparison with the critical value. The p-value will be evaluated from the calculated Z-value. Both the actions are based on the calculated test statistic value Z.

Step-by-step solution

Compute the Test Statistic

The first step is to compute the test statistic. This is done using the formula: \[ Z = \frac{{\bar{w} - μ_W}}{{\sqrt{s_W^{2}/n}}} \]where \(n\) is the number of observations. For the exercise, \(μ_W\) is zero, and given that \(\bar{w} = 2.07\), \(s_W^2 = 84.63\) and \(n = 51\), substituting these values in the formula gives:\[ Z = \frac{{2.07 - 0}}{{\sqrt{84.63 / 51}}} \]

Determine the Critical Value

Next, determine the critical value at the given significance level of \(α = 0.05\). There are tables available to find this value, frequently this value at 0.05 significance level is \(1.645\) for a one-tail test.

Decide Whether to Reject the Null Hypothesis

Now compare the test statistic (Z) with the critical value. If the test statistic is greater than the critical value, reject \(H_0\); otherwise, do not reject \(H_0\). Here, the null hypothesis will be rejected if \(Z > 1.645\).

Calculate the p-value

The p-value expresses the probability that the test statistic would have a value as extreme or more extreme than the one observed, assuming the null hypothesis is true. It's computed by looking up the test statistic Z in a standard normal distribution table, which is typically given in a statistics textbook. The p-value is 1 minus the probability \(Z\) under the standard normal distribution. If the p-value is less than or equal to the significance level, \(H_0\) is rejected; else, \(H_0\) is not rejected.

Key concepts

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to determine whether to reject the null hypothesis. For the exercise,

• we use the values: mean difference (\( \bar{w} \) = 2.07),
• sample variance (\( s_W^2 = 84.63 \)), and
• number of observations (\( n = 51 \)).

Substituting into the formula \[ Z = \frac{\bar{w} - \mu_W}{\sqrt{s_W^{2}/n}} \]where \( \mu_W = 0 \), gives a calculated test statistic \( Z \).
This value tells us how far away our sample mean is from the null hypothesis mean, measured in terms of standard deviations. In simpler words, it translates our sample results into a standard scale so we can compare it against theoretical values.
Understanding the test statistic helps determine the likelihood of observing the test data under the null hypothesis. This forms the basis for deciding whether the evidence is strong enough to reject the null hypothesis.

Significance Level

The significance level, denoted as \( \alpha \), is a threshold set by the researcher for rejecting the null hypothesis. In hypothesis testing, \( \alpha \) represents the probability of making a Type I error, which is rejecting a true null hypothesis.
Common significance levels include 0.05, 0.01, and 0.10, with \( \alpha = 0.05 \) being widely used. This means there's a 5% risk of concluding that a difference exists when there is none.
The chosen significance level affects the critical value, which is a point on the test distribution used to decide whether to reject \( H_0 \). In the current exercise, \( \alpha = 0.05 \) corresponds to a critical value of 1.645 for a one-tailed test.
By comparing the test statistic to this critical value, one determines whether the observed effect in the data is statistically significant.

P-value

The p-value is a crucial concept in hypothesis testing. It represents the probability of observing a test statistic as extreme, or more extreme, than the one observed, assuming the null hypothesis is true.
To determine the p-value, we look up the test statistic in a standard normal distribution table. The p-value tells us how likely our sample result would be if \( H_0 \) is true.
A smaller p-value indicates stronger evidence against \( H_0 \). If this p-value is less than or equal to the significance level \( \alpha \), we reject \( H_0 \). In our exercise, if the computed p-value is less than 0.05, the null hypothesis would be rejected.
This p-value approach allows us to quantify the strength of evidence against the null hypothesis, providing a straightforward method for decision-making.

Null Hypothesis

The null hypothesis, denoted as \( H_0 \), is a statement we make about a population parameter that we assume to be true until we have sufficient evidence to prove otherwise. In the context of the exercise, \( H_0: \mu_W = 0 \) implies no difference between the means of distances traveled by two brands of golf balls.
Testing \( H_0 \) involves computing a test statistic and comparing it against a critical value or using a p-value to reach a decision.
The aim is to assess whether the observed data provide enough statistical evidence to reject \( H_0 \) in favor of an alternative hypothesis, \( H_1 \).
Rejecting \( H_0 \) suggests that the observed effect is likely to be real and not due to random chance. This concept underscores all hypothesis tests, ensuring that our conclusions are built on a solid foundation of statistical reasoning.