Question

The manager at the Dixon’s seafood restaurant is interested in finding out if her servers earn similar tips during the lunch hour compared to the national average. Nationally, servers at diners earn $25 in the lunch hour with a standard deviation of $3.25. A random sample of size 4 servers were taken from the wait staff at the Dobbs Diner. The tips earned during the lunch hour were $18, $13, $27, and $23. Is the sample mean tips for wait staff at the Dobbs Diner significantly different from the national average of $25 during the lunch hour? (Hint: you will first need to calculate the sample mean and sample standard deviation using the data provided).
a. State both the null and alternative hypothesis in symbols.
b. Calculate the degrees of freedom (df) and find the t critical value with a significance level of .05.
c. Conduct the appropriate t-test to test your hypothesis (alpha = .05). Show all of your work.
d. Report your decision.
e.  Interpret your findings.

Answer 1

= (18 + 13 + 27+ 23)/4 = 20.25

s = sqrt(((18 - 20.25)^2 + (13 - 20.25)^2 + (27 - 20.25)^2 + (23 - 20.25)^2)/3) = 6.0759

a)

  

b) df = 4 - 1 = 3

At alpha = 0.05, the critical value is +/- t_0.025,3 = +/- 3.187

Reject H₀, if t < -3.187 or t > 3.187

c) The test statistic is

  

d) Since the test statistic value does not lie in the critical region, so we should not reject H₀.

e) At 0.05 significance level,there is not sufficient evidence to conclude that the sample mean tips for for wait staff at the Dobbs dinner is significantly different from the national average of $25 during the lunch hour.