Question

An experiment was conducted to compare the tenderness of meat cuts treated with two different meat tenderizers. Prior to applying the tenderizers, the data were paired by the specific meat cut from the same steer and by cooking paired cuts together. After cooking, each cut was rated by the same judge on a scale of \(1-10,\) with 10 corresponding to the most tender meat. Do the data provide sufficient evidence to indicate that one of the two tenderizers tends to receive higher ratings than the other? Would a Student's \(t\) -test be appropriate for analyzing these data? Explain. $$ \begin{array}{llr} \hline & {\text { Tenderizer }} \\ \text { Cut } & \text { A } & \text { B } \\ \hline \text { Shoulder roast } & 5 & 7 \\ \text { Chuck roast } & 6 & 5 \\ \text { Rib steak } & 8 & 9 \\ \text { Brisket } & 4 & 5 \\ \text { Club steak } & 9 & 9 \\ \text { Round steak } & 3 & 5 \\ \text { Rump roast } & 7 & 6 \\ \text { Sirloin steak } & 8 & 8 \\ \text { Sirloin tip steak } & 8 & 9 \\ \text { T-bone steak } & 9 & 10 \\ \hline \end{array} $$

Short answer

Is the Student's t-test appropriate for analyzing these data? Answer: No, there is insufficient evidence to indicate that one of the two tenderizers has a higher rating than the other. The Student's t-test is appropriate for analyzing these data.

Step-by-step solution

Calculate differences in ratings

Calculate the difference in tenderness ratings for each meat cut between Tenderizer A and Tenderizer B. Differences: [5-7, 6-5, 8-9, 4-5, 9-9, 3-5, 7-6, 8-8, 8-9, 9-10] = [-2, 1, -1, -1, 0, -2, 1, 0, -1, -1]

Calculate mean and standard deviation

Calculate the mean (µ) and standard deviation (σ) of the differences. Mean (µ): \((-2+1-1-1+0-2+1+0-1-1)/10 = -5/10 = -0.5\) Standard Deviation (σ): \(σ = \sqrt{\frac{(a_1-\mu)^2+(a_2-\mu)^2+\cdots+(a_{10}-\mu)^2}{n-1}} = \sqrt{\frac{(-2-(-0.5))^2+(1-(-0.5))^2+(\cdots)+(-1-(-0.5))^2}{10-1}} \approx 1.07\)

Calculate t-score

Calculate the t-score using the formula: \(t = \frac{\mu}{\frac{σ}{\sqrt{n}}}\) , where \(n\) is the sample size (amount of paired measurements). t = \(\frac{-0.5}{\frac{1.07}{\sqrt{10}}} \approx -1.57\)

Determine Degrees of Freedom and Critical Value

Determine the degrees of freedom (df) and two-tailed critical value (\(t_{critical}\)) for a chosen significance level (α). Let's use α = 0.05. Degrees of Freedom (df): \(n - 1 = 10 - 1 = 9\) Two-tailed critical value (\(t_{critical}\)) for α = 0.05 and df = 9: approximately \(\pm 2.262\)

Compare t-score with critical value and perform t-test

Compare the calculated t-score with the critical value and determine if the result is statistically significant. Since the calculated t-score (-1.57) falls within the two-tailed critical value range (-2.262, 2.262), we fail to reject the null hypothesis (H0). Based on the results of the paired samples t-test, there is insufficient evidence to indicate that one of the two tenderizers has a higher rating than the other. The Student's t-test is appropriate for analyzing these data.

Key concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make inferences about a population based on a sample of data. In the context of the paired sample t-test from the exercise, it involves following a systematic process to determine whether the observed differences in the meat tenderizer ratings are due to random chance or if they reflect a true effect.

Generally, we start by formulating two opposing hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1). The null hypothesis suggests that there is no effect or difference, while the alternative hypothesis suggests there is a significant effect or difference. For the meat tenderizer case, H0 would be that there is no difference in tenderness between the two tenderizers, and H1 would be that there is a difference.

After collecting and analyzing the data, we compute a test statistic that describes how much the sample data deviate from what we would expect under the null hypothesis. Finally, we make a decision about which hypothesis is supported by the sample data, operating within a pre-determined significance level to control the chance of making a Type I error—incorrectly rejecting the null hypothesis.

Statistical Significance

Statistical significance is a determination about the non-randomness of the results of a hypothesis test. When comparing the tenderizer ratings, reaching statistical significance means that it's unlikely the observed difference in ratings occurred by chance.

To assess this, a significance level (often denoted as \( \alpha \)) is chosen. The significance level represents the probability of making a Type I error and is typically set at 0.05, indicating a 5% risk of concluding there is an effect when there isn't one. If the test statistic falls within the critical region defined by this significance level, we reject the null hypothesis and conclude that the result is statistically significant, suggesting a real effect is present. In the tenderizer experiment, however, the result was not statistically significant, indicating that any difference in ratings could be due to chance.

Degrees of Freedom

Degrees of freedom (df) are a crucial component in the calculation of a t-test, representing the number of independent values in the data that are free to vary when estimating a statistical parameter.

In the paired sample t-test, the degrees of freedom are calculated as the number of paired observations minus one (df = n - 1). In our tenderizer example, there were 10 pairs of observations, so the degrees of freedom were 9. The degrees of freedom are used to determine the critical t-value from the t-distribution, which is a key part of the hypothesis testing. The critical t-value defines how far away the test statistic has to be from zero to reject the null hypothesis, considering the sample size and desired level of confidence.

T-Score Calculation

In a paired sample t-test, the t-score is a statistic that measures the relative difference between the paired observations compared to the variability of the differences. To calculate the t-score, we use the formula \( t = \frac{\mu}{\frac{\sigma}{\sqrt{n}}} \) where \( \mu \) is the mean of the differences, \( \sigma \) is the standard deviation, and \( n \) is the number of differences.

For the meat tenderizer experiment, we first computed the differences between the scores for each meat cut, then found the mean and standard deviation of these differences. Finally, we applied the t-score formula. The resulting t-score helps us to compare the observed mean difference to the expected mean difference under the null hypothesis. If the absolute value of the t-score is greater than the critical t-value found in the t-distribution table for our degrees of freedom, we reject the null hypothesis; otherwise, we fail to reject it—as was the case in this scenario with the tenderizer ratings.