Answer:
We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 30 minutes
Sample mean, [tex]\bar{x}[/tex] = 31.5 minutes
Sample size, n = 41
Alpha, α = 0.05
Sample standard deviation, s = 3.5 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{31.5 - 30}{\frac{3.5}{\sqrt{41}} } = 2.744[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.