Answer:
Using a significance level of 5% you can reject the null hypothesis, this means that the average time it takes to make a delivery from the local pizza chain isn't what the chain claims.
Explanation:
Hello!
Your study variable is X: time that takes to make a delivery from the local pizza chain.
X~N(μ; σ²)
μ= 30 min
σ= 10 min
The hypothesis is:
H₀: μ = 30
H₁: μ ≠ 30
α: 0,05
(There is no level of significance so I've choose one of the most common)
t= X[bar] - μ ~t
![_(n-1)](https://img.qammunity.org/2020/formulas/mathematics/college/4u2953x55uio4vma6787vmob44oyrl3bj3.png)
S/√n
The rejection region is two-tailed, the critical number are:
![t_(n-1;\alpha/2) = t_(19;0.025) = -2.093](https://img.qammunity.org/2020/formulas/mathematics/college/x2r1joqwtxealztbec9f9s0euq0kz48wth.png)
![t_(n-1;1-\alpha /2) = t_(19; 0.975) =2.093](https://img.qammunity.org/2020/formulas/mathematics/college/dgvggrbauphfm04ct9buhco1q4tv0dwu9k.png)
Sample:
n= 20
X[bar]= 42
S= 10
t= X[bar] - μ = 42 - 30 = 5.37
S/√n 10/√20
Using a significance level of 5% you can reject the null hypothesis, this means that the average time it takes to make a delivery from the local pizza chain isn't what the chain claims.
I hope it helps!