Question

The number of entrees purchased in a single order at a Noodles & Company restaurant has had a historical average of 1.45 entrees per order. On a particular Saturday afternoon, a random sample of 22 Noodles orders had a mean number of entrees equal to 1.85 with a standard deviation equal to 0.89. At the 5 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?

Answer 1

`t=(1.85-1.45)/((0.89)/(√(22)))=2.108`

`p_v =P(t_(21)>2.108)=0.0236`

If we compare the p value and a significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and the actual true mean for the entrees per order is not significantly higher than 1.45 minutes at 5% of significance.

Explanation:

Data given and notation

`\bar X=1.85` represent the sample mean

`s=0.89` represent the standard deviation for the sample

`n=22` sample size

`\mu_o =1.45` represent the value that we want to test

`\alpha=0.05` 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 to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 1.45 entrees per order, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 1.45`

Alternative hypothesis:
`\mu > 1.45`

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(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".

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

`t=(1.85-1.45)/((0.89)/(√(22)))=2.108`

Now we need to find the degrees of freedom for the t distirbution given by:

`df=n-1=22-1=21`

What do we can conclude?

Compute the p-value

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

`p_v =P(t_(21)>2.108)=0.0236`

If we compare the p value and a significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and the actual true mean for the entrees per order is not significantly higher than 1.45 minutes at 5% of significance.