Question

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: \({\bf{m = 50,}}\overline {\bf{x}} {\bf{ = 22}}{\bf{.63,}}{{\bf{s}}_{\bf{1}}}{\bf{ = 7}}{\bf{.82}}\)

No introduction: \({\bf{n = 50,}}\overline {\bf{y}} {\bf{ = 14}}{\bf{.15,}}{{\bf{s}}_{\bf{2}}}{\bf{ = 6}}{\bf{.10}}\)

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 \({\bf{\theta = }}{{\bf{\mu }}_{\bf{1}}}{\bf{ - 1}}{\bf{.5}}{{\bf{\mu }}_{\bf{2}}}\)

Short answer

Reject the null hypothesis and there is not sufficient evidence to support the claim that an introduction increases tips on average by more than 50%.

Step-by-step solution

Given information

\(\begin{array}{l}{{\bar x}_1} = 22.63,{s_1} = 7.82\\{{\bar x}_2} = 14.15 {s_2} = 6.10\\{n_1} = {n_2} = 50\end{array}\)

Since both sample sizes are large, we know that the sampling distributions of the sample mean are approximately normal (by the central limit theorem) and thus the statistic\({\bar X_1} - 1.5{\bar X_2}\)is approximately normally distributed.

Let us assume:\(\alpha = 0.05\)

Given claim: Increases by more than 50%.

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} - 1.5{\mu _2} = 0\\{H_a}:{\mu _1} - 1.5{\mu _2} > 0\end{array}\)

Find the test statistic

The value of the test statistic is the statistic decreased by its mean, divided by its standard deviation:

\(\begin{array}{c}z = \frac{{{{\bar x}_1} - 1.5{{\bar x}_2} - E\left( {{{\bar X}_1} - 1.5{{\bar X}_2}} \right)}}{{\sqrt {V\left( {{{\bar X}_1} - 1.5{{\bar X}_2}} \right)} }}\\ = \frac{{{{\bar x}_1} - 1.5{{\bar x}_2} - \left( {{\mu _1} - 1.5{\mu _2}} \right)}}{{\sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + 1.{5^2}\frac{{\sigma _2^2}}{{{n_2}}}} }}\\ = \frac{{22.63 - 1.5(14.15) - (0)}}{{\sqrt {\frac{{7.8{2^2}}}{{50}} + 1.{5^2}\frac{{6.1{0^2}}}{{50}}} }}\\ \approx

0.83\end{array}\)

Find the P-value

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z, assuming that the null hypothesis is true. Determine the probability using the normal probability table.

\(\)\(\begin{array}{c}P = P(Z > 0.83)\\ = 1 - P(Z < 0.83)\\ = 1 - 0.7967\\ = 0.2033\end{array}\)

If the P-value is smaller than the significance level\(\alpha \), then the null hypothesis is rejected.

\(P > 0.05 \Rightarrow {\rm{Fail to reject}} {H_0}\)

There is not sufficient evidence to support the claim that an introduction increases tips on average by more than 50%.

Final conclusion

There is not sufficient evidence to support the claim that an introduction increases tips on average by more than 50%.