Question

The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. On a particular Saturday afternoon, a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01. At the 2 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?

Answer 1

Yes, this sample show that the average number of entrees per order was greater than expected.

Explanation:

We are given that the number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. For this a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01.

We have to test that the average number of entrees per order was greater than expected or not.

Let, Null Hypothesis,
`H_0` :
`\mu` = 1.7 {means that the average number of entrees per order was same as expected of 1.7 entrees per order}

Alternate Hypothesis,
`H_1` :
`\mu` > 1.7 {means that the average number of entrees per order was greater than expected of 1.7 entrees per order}

The test statistics that will be used here is One sample t-test statistics;

T.S. =
`(Xbar-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where, Xbar = sample mean number of entrees = 2.1

s = sample standard deviation = 1.01

n = sample of Noodles = 48

So, test statistics =
`(2.1-1.7)/((1.01)/(√(48) ) )` ~
`t_4_7`

= 2.744

Now, at 2% significance level the critical value of t at 47 degree of freedom in t table is given as 2.148. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the average number of entrees per order was greater than expected.

Answer 2

`t=(2.1-1.7)/((1.01)/(√(48)))=2.744`

`p_v =P(t_((47))>2.744)=0.0043`

If we compare the p value and the significance level given
`\alpha=0.02` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

Explanation:

Data given and notation

`\bar X=2.1` represent the mean

`s=1.01` represent the sample standard deviation

`n=48` sample size

`\mu_o =1.7` represent the value that we want to test

`\alpha=0.02` 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 is higher than 1.7, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 1.7`

Alternative hypothesis:
`\mu > 1.7`

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=(\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".

Calculate the statistic

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

`t=(2.1-1.7)/((1.01)/(√(48)))=2.744`

P-value

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

`df=n-1=48-1=47`

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

`p_v =P(t_((47))>2.744)=0.0043`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.02` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.