Question

In an advertisement, a pizza shop claims its mean delivery time is 30 minutes. A consumer group believes that the mean delivery time is greater than the pizza shop claims. A random sample of 41 delivery times has a mean of 31.5 minutes with standard deviation of 3.5 minutes. Does this indicate at a 5% significance level that the mean delivery time is longer than what the pizza shop claims?

Answer 1

We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Explanation:

We are given the following in the question:

Population mean, μ = 30 minutes

Sample mean,
`\bar{x}` = 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

`H_(0): \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}`

We use one-tailed z test to perform this hypothesis.

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(31.5 - 30)/((3.5)/(√(41)) ) = 2.744`

Now,
`z_(critical) \text{ at 0.05 level of significance } = 1.64`

Since,

`z_(stat) > z_(critical)`

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.