91Ó°ÊÓ

Researchers gave 40 index cards to a waitress at an Italian restaurant in New Jersey. Before delivering the bill to each customer, the waitress randomly selected a card and wrote on the bill the same message that was printed on the index card. Twenty of the cards had the message, "The weather is supposed to be really good tomorrow. I hope you enjoy the day!" Another 20 cards contained the message, "The weather is supposed to be not so good tomorrow. I hope you enjoy the day anyway!" After the customers left, the waitress recorded the amount of the tip (percent of bill) before taxes. Here are the tips for those receiving the good-weather message: \({ }^{20}\) \(\begin{array}{llllllllll}20.8 & 18.7 & 19.9 & 20.6 & 21.9 & 23.4 & 22.8 & 24.9 & 22.2 & 20.3\end{array}\) \(\begin{array}{llllllllll}24.9 & 22.3 & 27.0 & 20.5 & 22.2 & 24.0 & 21.2 & 22.1 & 22.0 & 22.7\end{array}\) The tips for the 20 customers who received the bad weather message are \(18.0 \quad 19.1 \quad 19.2 \quad 18.8 \quad 18.4 \quad 19.0 \quad 18.5 \quad 16.1 \quad 16.8 \quad 14.0\) \(\begin{array}{lllllllll}17.0 & 13.6 & 17.5 & 20.0 & 20.2 & 18.8 & 18.0 & 23.2 & 18.2\end{array}\) (a) Make stemplots or histograms of both sets of data. Because the distributions are reasonably symmetric with no extreme outliers, the \(t\) procedures will work well. (b) Is there good evidence that the two different messages produce different percent tips? State hypotheses, carry out a two-sample \(t\) test, and report your conclusions.

Step by step solution

01

Create Stem-and-Leaf Plots

Start by arranging the data into two separate stem-and-leaf plots. For the `good weather` message tips, list each value with its stem as tens and leaves as tenths. Do the same for the `bad weather` message tips. This visual representation will help us see the distribution of the data.

02

Calculate Means and Standard Deviations

Calculate the mean (\bar{x}) for each group by adding all the values and dividing by the number of values (20 in each group). Then, calculate the standard deviation (s) using the formula \( s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \), where \( x_i \) are individual data points.

03

Formulate Hypotheses

State the null hypothesis (H_0) and the alternative hypothesis (H_a). Here, \( H_0: \mu_1 = \mu_2 \) (no difference in mean tips) and \( H_a: \mu_1 eq \mu_2 \) (a difference in mean tips), where \( \mu_1 \) and \( \mu_2 \) are the means of the two groups.

04

Conduct Two-Sample t-Test

Use the two-sample t-test formula \( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }} \). Calculate \( t \) by substituting mean and standard deviation values from Step 2. Determine degrees of freedom using \( df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{ \left( \frac{s_1^2}{n_1} \right)^2 }{n_1-1} + \frac{ \left( \frac{s_2^2}{n_2} \right)^2 }{n_2-1} } \).

05

Compare t-score to Critical t, Draw Conclusions

Find the critical t-value from the t-distribution table at your chosen significance level (typically 0.05/5%) and the calculated degrees of freedom. Compare the computed t-score from Step 4 to this critical t-value. If \( t \) is beyond the critical t-value, reject \( H_0 \). Otherwise, do not reject \( H_0 \). Conclude if the message has statistically significant effect on tips.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stem-and-Leaf Plot

A Stem-and-Leaf plot is a simple yet effective way to visualize numerical data. It's similar to a histogram but retains the original data values. To create a stem-and-leaf plot, divide each data value into a `stem` and a `leaf`. Typically, the stem is all but the final digit, and the final digit becomes the leaf. For instance, in our exercise, you might represent a tip of 22.3 as a stem of 22 and a leaf of 3.

• This method helps to see the shape of the dataset's distribution and detect any potential outliers.
• It also allows a quick glance to determine if data should be considered symmetric or skewed.

Breaking down the tips from the good weather message into a stem-and-leaf plot reveals a clear summary of the data’s distribution. Similarly, you'll do this for tips from the bad weather group, allowing easy comparison between the two sets.

Hypothesis Testing

Hypothesis testing is a statistical method to make decisions about populations based on sample data. In the context of our exercise, we are checking if different weather messages influence tipping behavior.Here's how it works:

• Null Hypothesis (H₀): This states that there is no effect or difference. In our exercise, this means the message has no impact on the tip percentage: \( \mu_1 = \mu_2 \).
• Alternative Hypothesis (H_a): This opposes the null. It suggests there is a meaningful difference: \( \mu_1 eq \mu_2 \).

The process involves using statistical tests to decide whether to reject the null hypothesis. Typically, you'll employ a statistical test such as a t-test when the data are approximately normally distributed and the sample size is small, as is the case here.

Statistical Significance

Statistical significance tells us whether the results of an experiment are likely not due to chance. It's calculated after performing tests like the two-sample t-test. In our exercise, once the t-score is calculated:

• Find the critical t-value from tables for a chosen significance level, usually 0.05, which corresponds to a 5% chance of wrongly rejecting the null hypothesis when it is true.
• Compare the calculated t-score with the critical value. If the t-score is higher than the critical value, we reject the null hypothesis, indicating that the difference in average tips due to the message is statistically significant.

This concept is crucial as it helps us conclude if the differences or effects studied in the data are likely "real" or simply occurred by random chance. If the difference is statistically significant, action or further investigation might be warranted.