Suppose we would like to determine if the average spent per customer for dinner

Question

Suppose we would like to determine if the average spent per customer for dinner at a new restaurant in town is more than $20.00. A sample of 49 customers over a three-week period was randomly selected and the average amount spent was $22.00. Assume that the standard deviation is known to be $5.50. Using a 95% level of significance, would we conclude the average amount spent per customer is more than $20.00? What is the statements of this hypothesis testing? What is the value of alpha, i.e. the probability of the type one error? What is the test statistic of this hypothesis testing? what is the equation to calculate the value of this test statistic? and what is the observed value of this test statistic in this sample? What is the critical region and what is the p value. What is your conclusion on the hypothesis testing, i.e. reject H_0 of don't reject and why? What is the scientific conclusion?

Explanation / Answer

Question 1

The null and alternative hypothesis for the given test is given as below:

Null hypothesis: H0: The average amount spent per customer for dinner at a new restaurant in town is $20.00.

Alternative hypothesis: Ha: The average amount spent per customer for dinner at a new restaurant in town is more than $20.00.

H0: µ = 20 versus Ha: µ > 20

This is a one tailed test. This is a right tailed test. This is a upper tailed test.

Question 2

The value of alpha or the probability of type one error is given as 1 – c = 1 – 0.95 = 0.05 or 5%.

Question 3

Here we have to use the one sample Z test for the population mean. The formula for test statistic Z is given as below:

Test statistic = Z = (Xbar - µ) / [/sqrt(n)]

Where n is the sample size.

We are given

Xbar = 22.6

µ = 20

= 5.5

n = 49

Z = (22.6 – 20) / [5.5/sqrt(49)]

Z = 3.3091

Observed value of test statistic is given as Z = 3.3091.

Question 4

The critical region and p-value is given as below:

Critical value = 1.6449

Critical region = Z > 1.6449

P-value = 0.0005

Question 5

For the given test, we get p-value = 0.0005 which is less than alpha value 0.05, so we reject the null hypothesis H0.

Question 6

We conclude that there is sufficient evidence that the average amount spent per customer for dinner at a new restaurant in town is more than $20.00.

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