Question

Weights of Running Shoes The weights in ounces of a sample of running shoes for men and women are shown. Test the claim that the means are different. Use the \(P\) -value method with \(\alpha=0.05 .\) $$ \begin{array}{ll|ccc}{\text { Men }} & {} & {\text { Women }} & {} \\ \hline 10.4 & {12.6} & {10.6} & {10.2} & {8.8} \\ {1.1} & {14.7} & {9.6} & {9.5} & {9.5} \\ {10.8} & {12.9} & {10.1} & {11.2} & {9.3} \\ {11.7} & {13.3} & {9.4} & {10.3} & {9.5} \\ {12.8} & {14.5} & {9.8} & {10.3} & {11.0} \\\ \hline\end{array} $$

Short answer

The means of the weights are likely different if the P-value is less than 0.05; otherwise, they are not significantly different.

Step-by-step solution

Gather the Data

We start by listing down the weights of the running shoes for men and women from the table. For men: \([10.4, 12.6, 10.6, 10.2, 8.8, 1.1, 14.7]\). For women: \([9.6, 9.5, 9.5, 10.8, 10.1, 11.2, 9.3, 9.4, 10.3, 9.5, 10.3, 11.0]\).

State the Hypotheses

The null hypothesis \(H_0\) is that the means of the weights of shoes for men and women are equal: \(H_0: \mu_1 = \mu_2\). The alternative hypothesis \(H_a\) is that the means are different: \(H_a: \mu_1 eq \mu_2\).

Calculate Means and Standard Deviations

Calculate the mean weight for men's shoes \(\bar{x}_1 = \frac{ \sum x_1}{n_1}\) and for women's shoes \(\bar{x}_2 = \frac{ \sum x_2}{n_2}\), where \(n_1\) and \(n_2\) are the sample sizes. Then, calculate the standard deviations \(s_1\) and \(s_2\). For men: \(n_1 = 7, \bar{x}_1 \approx 11.9, s_1 \approx 4.3\). For women: \(n_2 = 12, \bar{x}_2 \approx 9.8, s_2 \approx 0.78\).

Perform Independent T-Test

The formula for the t-statistic is \(t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\). Substituting the values, we calculate the t-statistic.

Determine Degrees of Freedom

Degrees of freedom (df) can be approximated using the formula for unequal variances: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \] Use this to find the appropriate critical t-value.

Compute the P-Value

Using a t-distribution table or software, find the P-value based on the computed t-statistic and degrees of freedom. The P-value helps in deciding whether to reject the null hypothesis.

Compare P-Value with Alpha

If the P-value \(< \alpha = 0.05\), we reject the null hypothesis. If the P-value \(\geq 0.05\), we fail to reject the null hypothesis.

Key concepts

Independent T-Test

The Independent T-Test is a vital statistical tool used to determine if there is a significant difference between the means of two independent groups. In this case, we are comparing the weights of men's and women's running shoes.
Unlike paired samples, here the samples are independent of each other, meaning that observations in one group do not affect those in the other group.
This test assumes that the data is normally distributed and that the variances of the two groups are equal or approximately equal.
To perform the test:

• Calculate the mean and standard deviations of both groups.
• Use these values to calculate the t-statistic.

The t-statistic helps in comparing the observed data against the null hypothesis.

P-Value Method

The P-Value Method is central to hypothesis testing. It assists in deciding whether to reject the null hypothesis. A P-value measures the strength of evidence against the null hypothesis provided by the data.
If the P-value is small, typically less than the significance level \( \alpha \), it indicates that the observed data is unlikely under the null hypothesis.
In most cases, \( \alpha \) is set at 0.05:

• If the P-value < \( \alpha = 0.05 \), we reject the null hypothesis, concluding there is a significant difference.
• If the P-value \( \geq 0.05 \), we fail to reject it.

The P-value, obtained from a t-distribution table or software, reflects the probability of obtaining test results at least as extreme as the ones observed.

Null Hypothesis

The Null Hypothesis (H_0) is a statement used in statistical analysis that proposes no significant difference exists between the means of two groups.
In our exercise, the null hypothesis is: H_0:  \(mu_1 = mu_2\), meaning the average weights of the shoes for men and women are the same.
This hypothesis serves as the starting assumption for hypothesis testing. It represents the case of no effect or no difference.
The aim of testing is to evaluate its validity given the sample data:

• If evidence suggests a significant difference, we reject H_0.
• If not, we maintain H_0 as a plausible explanation.

The decision to reject or not reject relies heavily on the P-value and the predetermined significance level.

Degrees of Freedom

Degrees of Freedom (df) is a statistical concept that reflects the number of independent values or quantities that can vary in an analysis without violating any constraints.
For an Independent T-Test with unequal variances, df is approximated as:
\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]
Here, \( s_1 \) and \( s_2 \) are the standard deviations of the two samples, and \( n_1 \) and \( n_2 \) are their respective numbers of observations.
The degrees of freedom play a critical role in determining the critical t-value needed to assess the significance of the test statistic against the null hypothesis.
By understanding df, you can ensure you are interpreting the test results accurately, especially in complex scenarios where variances are not equal.