verification of work

Hi, looking for a verification of my answers for this questions: And if I did it wrong please tell me how to improve. Thanks!

Rick, the new manager of a trendy restaurant in Ottawa, has many questions about his customers and their preferences for dinner-time beverages. He carries out a random survey of 300 customers and obtains this two-way table of observed frequencies.

	Water	Soft Drink	Beer	Total
21-34 years	30	90	20	140
35-55 years	40	40	20	100
Over 55 years	20	10	30	60
Total	90	140	70	300

Rick’s main question is about whether age and beverage preference within the restaurant’s customer population are related.
a) What is the name of the appropriate hypothesis test that should be used to answer this?
b) To carry out this test, fill in the following table of expected frequencies. Fill in the blanks in the “Water” column.

	Water	Soft Drink	Beer	Total
21-34 years		65.3	32.7	140
35-55 years		46.7	23.3	100
Over 55 years		28.0	14.0	60
Total	90	140	70	300

c) State clearly the hypotheses to be tested.
Complete the test, using a 5% significance level, by determining the test statistic, giving the degrees of freedom, the critical value OR upper bound on a P-value, and a conclusion. You may use Minitab to calculate the test statistic but demonstrate you know how to calculate it manually by providing the manual calculations for the contribution of the cell "Water - 21-23 years" to the test statistic.

Answers:

a) The appropriate test to use is the chi-square test of independence.

b)

	Water	Soft Drink	Beer	Total
21-34 years	42	65.3	32.7	140
35-55 years	13	46.7	23.3	100
Over 55 years	18	28.0	14.0	60
Total	90	140	70	300

c)

H0= Customer age and their beverage preference unrelated (independent)
Ha= Customer age and their beverage preference are related (dependent)

d)

x^2 = sum (Observed – Expected)^2 / Expected

(30-42)^2 / 42 = 3.43

(90-65.3)^2 / 65.3 = 9.34

(20-32.7)^2 / 32.7 = 4.93

(40-30)^2 / 30 = 3.33

(40-46.7)^2 / 46.7 = 0.96

(20-23.3)^2 / 23.3 = 0.47

(20-18)^2 / 18 = 0.22

(10-28)^2 / 28 = 11.57

(30-14)^2 / 14 = 18.27

Summing the above, x^2 = 52.52

Degrees of freedom (Df) = (3-1)(3-1)=4
The corresponding p-value for a test statistics of 52.52 with 4 degrees of freedom is approximately 0. This means we reject the null since 0 is < our alpha value of 0.05.

Hey dumbledore.

Your working looks OK, but as a suggestion for the future: you should probably post your data so that it can be evaluated in some kind of statistical package like R.

The reason is that checking all the calculations by hand are tedious and are prone to errors.

For this reason if you say posted the outputs and inputs of data into a package, it's much easier to verify since all we have to do is check the inputs and run the code on our machine to verify since the package is going to be the same and do the same thing.

I suspect that your results are right, but the only question I have is if you chose the independent contingency table to have a uniform distribution with respect to the frequencies in the table (I suspect you did this).

If you do the above, then it will be easier for other people to check and I guarantee that more people will want to check it since it will be easier on them to do so.