Question

To compare the demand for two different entrees, the manager of a cafeteria recorded the number of purchases of each entree on seven consecutive days. The data are shown in the table. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the Excel printout. $$ \begin{array}{lcc} \text { Day } & \mathrm{A} & \mathrm{B} \\ \hline \text { Monday } & 420 & 391 \\ \text { Tuesday } & 374 & 343 \\ \text { Wednesday } & 434 & 469 \\ \text { Thursday } & 395 & 412 \\ \text { Friday } & 637 & 538 \\ \text { Saturday } & 594 & 521 \\ \text { Sunday } & 679 & 625 \end{array} $$

Short answer

Provide a short answer with reference to the p-value and the significance level. Answer: Since the p-value (0.627) is greater than the chosen significance level (0.05), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that there is a significant difference in mean demand between the two entrees.

Step-by-step solution

State the hypotheses

We will test the following hypotheses: $$ H_0: \mu_A = \mu_B \\ H_a: \mu_A \neq \mu_B $$ Where \(\mu_A\) and \(\mu_B\) are the population means of the demand for entree A and entree B, respectively. The null hypothesis, \(H_0\), states that there is no significant difference between the mean demands of the two entrees. The alternative hypothesis, \(H_a\), states that there is a significant difference in the mean demands of the two entrees.

Find the sample means and standard deviations

Based on the provided data, calculate the sample means and standard deviations for the two samples: Entree A: - Mean: \(\bar{x}_A = \frac{420+374+434+395+637+594+679}{7} = 504.86\) - Standard deviation: \(s_A \approx 137.21\) Entree B: - Mean: \(\bar{x}_B = \frac{391+343+469+412+538+521+625}{7} = 471.29\) - Standard deviation: \(s_B \approx 97.01\)

Choose the appropriate t-test

In this case, we will use the independent samples t-test since we are comparing the means of two different groups (entree A and entree B) and we do not know the population standard deviations.

Perform the t-test

Using the sample data, calculate the t-statistic and degrees of freedom: - Pooled variance: \(s_p^2 = \frac{(n_A - 1) s_A^2 + (n_B - 1) s_B^2}{n_A + n_B - 2} = \frac{6\cdot 137.21^2 + 6 \cdot 97.01^2}{12} \approx 14542.29\) - Standard error: \(SE=\sqrt{\frac{s_p^2}{n_A}+\frac{s_p^2}{n_B}}=\sqrt{\frac{14542.29}{7}+\frac{14542.29}{7}}\approx 66.75\) - t-statistic: \(t=\frac{\bar{x}_A-\bar{x}_B}{SE}=\frac{504.86-471.29}{66.75}\approx 0.50\) - Degrees of freedom: \(df=n_A+n_B-2=7+7-2=12\)

Calculate the p-value

Using a t-distribution table or software/calculator, find the p-value associated with the calculated t-statistic and degrees of freedom. Since this is a two-tailed test, we need to consider both tails of the distribution: $$ p = 2 \cdot P(T > 0.50) \approx 0.627 $$

Compare the p-value with the significance level

If the calculated p-value is less than our chosen significance level (commonly \(\alpha=0.05\)), we reject the null hypothesis. In this case the p-value (\(0.627\)) is greater than the significance level, so we fail to reject the null hypothesis.

Conclude the test

Since we fail to reject the null hypothesis, we do not have sufficient evidence to conclude that there is a significant difference in mean demand between the two entrees.

Key concepts

t-test

A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is particularly useful when dealing with small sample sizes, as it accounts for sample variability.
In this case, we are using an independent samples t-test to compare the mean demands of two entrees over a week. This means we are examining whether there is a difference in mean purchases of entree A and entree B, without assuming that the data is paired or related in any way.
The t-test involves calculating a t-statistic, which measures the size of the difference relative to the variation in the sample data. To find the t-statistic, we calculate the pooled variance, standard error, and degrees of freedom, allowing us to see if the observed differences could reasonably occur under the assumption of the null hypothesis.

null hypothesis

The null hypothesis, often denoted as \(H_0\), is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference, serving as the default position we aim to test against.
In the context of the exercise, the null hypothesis \(H_0: \mu_A = \mu_B\) suggests that there is no significant difference in the mean demand between entree A and entree B over the course of seven days.
This hypothesis assumes that any observed differences in the sample means could purely be due to random chance. The goal of hypothesis testing is to determine whether the evidence from the data is strong enough to reject this assumption.

• It acts as a starting point for statistical testing.
• Rejection of the null hypothesis suggests evidence against it.

alternative hypothesis

The alternative hypothesis, denoted by \(H_a\), is the statement that we are looking to provide evidence for in hypothesis testing. It usually represents a new effect or difference that is contrary to the null hypothesis.
In the analysis, the alternative hypothesis \(H_a: \mu_A eq \mu_B\) proposes that there is a significant difference in the mean demand of the two entrees.
This means we want to check if entree A has a noticeably different average number of purchases compared to entree B. If the p-value obtained is small enough, it suggests that \(H_0\) is unlikely to be true, thus supporting \(H_a\).

• \(H_a\) is what researchers typically aim to prove.
• A significant result indicates that the observed difference is unlikely under \(H_0\).

p-value

The p-value is a critical component in the process of hypothesis testing, as it helps determine the statistical significance of the test. It represents the probability of obtaining the observed data, or something more extreme, assuming that the null hypothesis is true.
In this scenario, the p-value computed is approximately 0.627. A p-value higher than the common significance level of 0.05 means we do not have enough evidence to reject the null hypothesis \(H_0\).
This large p-value suggests that the differences in mean purchases between entree A and entree B can be explained by random variation, rather than a genuine difference in demand.

• Low p-values indicate stronger evidence against \(H_0\).
• In this case, a p-value greater than 0.05 leads us to "fail to reject" \(H_0\).