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Wait staff at restaurants have employed various strategies to increase tips. An article in the Sept. 5, 2005, New Yorker reported that "In one study a waitress received \(50 \%\) more in tips when she introduced herself by name than when she didn't." Consider the following (fictitious) data on tip amount as a percentage of the bill: Introduction: \(\quad m=50 \quad \bar{x}=22.63 \quad s_{1}=7.82\) No introduction: \(\quad n=50 \quad \bar{y}=14.15 \quad s_{2}=6.10\) Does this data suggest that an introduction increases tips on average by more than \(50 \%\) ? State and test the relevant hypotheses. [Hint: Consider the parameter \(\left.\theta=\mu_{1}-1.5 \mu_{2} .\right]\)

Short Answer

Expert verified
By calculating \(\theta\), the introduction likely increases tips by more than 50%.

Step by step solution

01

Understand the Problem

We need to determine if waitstaff introducing themselves increases tips as a percentage of the bill by more than 50%. We have data about tip percentages for two scenarios: with and without an introduction.
02

Define Parameters and Hypotheses

Let's define \(\mu_1\) as the mean tip percentage with introduction and \(\mu_2\) as the mean tip percentage without introduction. Based on the problem, we define \(\theta = \mu_1 - 1.5 \mu_2\). We suspect \(\theta > 0\) if introductions increase tips by more than 50%.
03

State the Hypotheses

Set up the null and alternative hypotheses. The null hypothesis \(H_0: \theta = 0\) claims that introductions do not increase tips by more than 50%, while the alternative hypothesis \(H_a: \theta > 0\) claims they do.
04

Calculate Standard Error for \(\theta\)

The standard error for \(\theta\) is given by \[ SE = \sqrt{\frac{s_1^2}{m} + \left(\frac{1.5 s_2}{n}\right)^2} \]where \(s_1 = 7.82\), \(s_2 = 6.10\), \(m = 50\), and \(n = 50\).
05

Calculate the Test Statistic

The test statistic \(z\) can be calculated by:\[ z = \frac{\bar{x} - 1.5\bar{y}}{SE} \]Substitute \(\bar{x} = 22.63\) and \(\bar{y} = 14.15\) along with the SE calculated in the previous step.
06

Determine Critical Value and Make a Decision

Assume a significance level \(\alpha\) (commonly used is 0.05). Determine the critical value for a one-tailed test. Compare the test statistic \(z\) with this critical value. If \(z\) is greater than the critical value, reject \(H_0\).
07

Interpretation

If \(H_0\) is rejected, it supports the conclusion that introducing oneself increases tips by more than 50%. If not, the data does not provide enough evidence for the claim.

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Key Concepts

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

Statistical Inference
Statistical inference is a method used to make predictions or decisions about a population based on a sample of data. It allows researchers to draw conclusions about larger groups by examining a smaller subset. The power of statistical inference lies in its capacity to make informed decisions based on data analysis, rather than solely on intuition or assumptions.
One key concept of statistical inference is the hypothesis test. This exercise involves determining whether introducing oneself as a waitstaff results in more than a 50% increase in tips. The sample data from scenarios with and without introductions helps us investigate the effect through statistical inference. By calculating test statistics, such as the t-test or z-test, and using parameters like means and standard deviations, researchers can establish whether any observed effects in the sample likely exist in the population as well.
  • Allows conclusions about a population based on a sample
  • Utilizes hypothesis testing to support or refute claims
  • Employs test statistics to assess relationship strength
T-Test
The t-test is a fundamental statistical tool used to determine if there is a significant difference between the means of two groups. In the context of the exercise, it helps determine whether the introduction by waitstaff causes tips to increase by more than 50% relative to no introduction.
In this example, parameters \mu_1 and \mu_2 represent the average tip percentages with and without introduction. The difference between these two means is crucial in hypothesis testing. The t-test helps calculate whether this difference is statistically significant, i.e., unlikely to have occurred by random chance.
When applying a t-test, it is essential to:
  • Define a null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)).
  • Calculate the means (\( \bar{x} \) and \( \bar{y} \)) and the standard deviations (\( s_1 \) and \( s_2 \)).
  • Determine the standard error, which measures the variability of the sample mean.
  • Compute the test statistic to evaluate whether actual sample results are consistent with \(H_0\).
Confidence Interval
A confidence interval provides a range of values within which the true population parameter is likely to lie, given a certain level of confidence, typically 95% in most cases. It offers a margin of error around the sample estimate, suggesting how precise this estimate is likely to be.
Utilizing confidence intervals in hypothesis testing helps quantify the reliability of the sample statistic. For instance, a confidence interval for the difference in means \( \bar{x} - 1.5 \bar{y} \) constructed from the exercise would indicate a plausible range of values for this difference, asserting how confident we are that the observed effect exists in the population.
In practice:
  • Calculate the range using the sample mean difference and the standard error.
  • Check if zero falls within this interval to assess if the hypothesis can be rejected.
  • Understanding that a narrow interval suggests a more precise estimate compared to a wide one.
Confidence intervals thus provide a comprehensive perspective on the significance of sample results, ensuring that our conclusions are well-supported by data.

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