Question

A pizza delivery chain advertises that it will deliver your pizza in no more 20 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than advertised. For the simple random sample of 63 customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 20.49 minutes with a standard deviation of 1.42 minutes. Perform a hypothesis test using a 0.01 level of significance.

Answer 1

Answer:

$t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738$    

$p_v =P(t_{(62)}>2.738)=0.0040$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .  

Step-by-step explanation:

Data given and notation  

$\bar X=20.49$ represent the mean time for the sample  

$s=1.42$ represent the sample standard deviation for the sample  

$n=63$ sample size  

$\mu_o =20$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean time is actually higher than 20 min, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 20$  

Alternative hypothesis:$\mu > 20$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=63-1=62$  

Since is a one side right tailed test the p value would be:  

$p_v =P(t_{(62)}>2.738)=0.0040$  

And we can use the following excel code to find it:

"=1-T.DIST(2.738,62,TRUE)"

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .