Skip to main content

Section 25.2 Investigation 5.5: Restaurant Spending and Music

Exercises 25.2.1 The Study

A British study (North, Shilcock, & Hargreaves, 2003) examined whether the type of background music playing in a restaurant affected the amount of money that diners spent on their meals. The researchers asked a restaurant to alternate classical music, popular music, and silence on successive nights over 18 days.
Restaurant setting for the Investigation 5.5 study context
The following summary statistics about the total bills were reported:
Classical music Pop music No music
Mean 24.13 21.91 21.70
SD 2.243 2.627 3.332
Sample size \(n_1=120\) \(n_2=142\) \(n_3=131\)

1. State Hypotheses.

State the null and alternative hypotheses corresponding to the researchers’ conjecture (in symbols and/or in words, be sure to define any symbols you use).
Solution.
\(H_0\text{:}\) the true treatment means \((\mu_{class} = \mu_{pop} = \mu_{none})\) are all equal.
Or \(H_0\text{:}\) the mean amounts spent in the long run are the same regardless of type of music being played.
\(H_a\text{:}\) at least one true treatment mean differs.
Or \(H_a\text{:}\) at least one type of music shows a tendency for higher or lower spending amounts on average.

2. Calculate Key ANOVA Quantities.

Based on these summary statistics, calculate:
  • the overall (weighted) mean amount spent, \(\bar{x} = \sum (n_i \bar{x}_i) / \sum n_i\text{,}\)
  • the between-group variation, \(MSG = \sum n_i (\bar{x}_i - \bar{x})^2 / (I-1)\text{,}\)
  • and the pooled standard deviation, \(MSE = \sum_{i = 1}^I (n_i - 1)s_i^2 / (n - I)\text{.}\)
Check the consistency of your calculations with the summary statistics reported above.
Solution.
weighted mean = \([120(24.13) + 142(21.91) + 131(21.70)]/(120+142+131) = 22.52\) (this is in the "middle" of the 3 observed averages).
Between group variability.
Pooled variance = \([119(2.243^2)+141(2.627^2)+130(3.332^2)]/(119+141+130) = 7.73\text{.}\)
Pooled std dev = \(\sqrt{7.73} = 2.78\) (this is in the "middle" of the observed standard deviations).
variability between groups = \(120(24.13-22.52)^2 + 142(21.91-22.52)^2 + 132(21.7-22.52)^2/2 = 226\text{.}\)

Technology Detour.

3. ANOVA Calculations.
Confirm your calculations using technology.
R output for Investigation 5.5 technology detour
JMP output for Investigation 5.5 technology detour
Hint 1. Applet Instructions
Enter your observed \(F\) value and use the upper-tail probability.
Hint 2. R Instructions
Use pf(x, df1, df2, lower.tail=FALSE) with x = 29.2, df1 = 2, and df2 = 390.
Hint 3. JMP Instructions
Use Distribution Calculator, choose F, and compute the upper-tail probability for \(F = 29.2\) with df 2 and 390.
Hint 4. Solution
You should obtain F = 226/7.73 = 29.2. The p-value is approximately zero.

4. Interpret Relative Variation.

Which appears to be larger, the variation in amount spent between diners or the variation in amount spent between types of music?
Solution.
The variation between types of music appears larger than the variation in amount spent between diners. (MSGroups is much larger than MSError).

5. Assess ANOVA Conditions.

Do you consider this p-value to be valid? In other words, are the technical conditions for ANOVA met for this study design?
Solution.
We would need to be able to verify the technical conditions (in fact, there is an issue here in that the treatments were assigned to the evenings and not the individual dinners).

Study Conclusions.

In this study, a larger average amount was spent while classical music was playing. With such a large test statistic and such a small p-value (F = 31.48, p-value \(\approx\) 0), we have very strong evidence that the observed differences in the sample means could not have arisen by chance alone. However, we need to assess the technical conditions before we could know whether this inferential procedure is valid with these data. Although the equal standard deviation assumption seems reasonable (because 3.332/2.243 < 2), we do not have the sample data to examine the shapes of the sample distributions. Still, the sample sizes are large and the F procedure is fairly robust to departures from the normality condition. However, we should not consider these observations to be independent within each group, because the treatments were assigned to the evenings, not to the individual diners. In this sense, we only have a few replications per evening and ANOVA is not the appropriate analysis. Descriptively it appears that the average amount spent is larger when classical music is playing, but because of the confounding with evening, it’s possible that classical music was played on certain days of the week and that that led to more spending. This was a randomized experiment, so if it weren’t for the concern about independence, we would have been able to draw a cause-and-effect conclusion from the type of music played and the amount spent. We would want to be very cautious in generalizing these results to other restaurants and even other times of the year.

Subsection 25.2.2 Practice Problem 5.5A

Another way to analyze the data for the trial of Dr. Spock is to look at the percentages of women on the different venires and determine whether the mean percentage of women is equal across the seven judges. Below are the percentages of women on the venires for a recent sample from each of the judges. These data are below and in SpockPers.txt.
Judge 1 Judge 2 Judge 3 Judge 4 Judge 5 Judge 6 Judge 7
16.8 27.0 21.0 24.3 17.7 16.5 6.4
30.8 28.9 23.4 29.7 19.7 20.7 8.7
33.6 32.0 27.5 21.5 23.5 13.3
40.5 32.7 27.5 27.9 26.4 13.6
48.9 35.5 30.5 34.8 26.7 15.0
45.6 31.9 40.2 29.5 17.7
32.5 29.8 18.6
33.8 31.9 23.1
33.8 36.2 15.2

Checkpoint 25.2.2. What New Information Does ANOVA Add?

Explain what information we learn from analyzing the data this way that we did not see when we carried out the Chi-squared test on the overall proportion of women for each judge. Why might this information be useful?

Checkpoint 25.2.3. Summarize Judge Percentages.

Produce numerical and graphical summaries to compare the percentages across the seven judges.

Checkpoint 25.2.4. Run ANOVA on Judge Percentages.

Carry out an ANOVA to test whether at least one judge has a different mean percentage. Did you state the null and alternative hypotheses in terms of population parameters or in terms of treatment effects?

Checkpoint 25.2.5. Check Technical Conditions.

Comment on whether you believe the technical conditions for this procedure are met.

Subsection 25.2.3 Practice Problem 5.5B

Recall the Disability Discrimination study (Investigation 5.4). Suppose the researchers had been most interested in comparing the results for those in a wheelchair to those with leg amputation.

Checkpoint 25.2.6. Two-sample \(t\)-test.

Carry out a two-sided two-sample pooled t-test (see Investigation 4.2: use the technology option that assumes the variances are equal) to assess whether there is a statistically significant difference in the average ratings assigned to these two groups.

Checkpoint 25.2.7. ANOVA for the Same Two Groups.

Carry out an analysis of variance to assess whether there is a statistically significant difference in the average ratings assigned to these two groups.

Checkpoint 25.2.8. Relate the \(t\) and \(F\) Results.

How are the p-values from the two approaches related? How do you think the t-test statistic and the F-test statistic are related? Explain.

Checkpoint 25.2.9. When to Use \(t\) vs ANOVA.

Suggest, in general, a situation where the two-sample t-procedure would be preferred and a situation where the ANOVA procedure would be preferred. [Hint: What would be true about the research question?]
You have attempted of activities on this page.