Question

You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $12 (μ = 42, σ = 12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the α = 0.05 level of significance.

Answer 1

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

H₀: µ = 42 versus H_a: µ ≠ 42

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 42

σ = 12

n = 16

Xbar = 44.50

α = 0.05

Critical value = 1.96

(by using z-table or excel)

Z = (44.50 – 42)/[12/sqrt(16)]

Z = 0.8333

P-value = 0.4047

(by using Z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the server make an amount of $42 on an average.