Question

A distributor of personal computers has five locations in the city. The sales in units for the first quarter of the year were as follows:

Location	Observed Sales (units)
North Side	70
Pleasant Township	75
Southwyck	70
I-90	50
Venice Avenue	35

At the .01 level of significance, what is your decision?

Select one:

a.

Reject the null hypothesis because the computed p-value is less than the level of significance.

b.

Reject the null hypothesis because the computed p-value is greater than the level of significance.

c.

Do not reject the null hypothesis because the computed p-value is greater than the level of significance.

d.

Do not reject the null hypothesis because the computed p-value is less than the level of significance.

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: Data follows the given distribution.

Alternative hypothesis: H_a: Data do not follow the given distribution.

We assume/given level of significance = α = 0.01

We are given

Number of categories = N = 5

Degrees of freedom = df = N - 1 = 4

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation tables for test statistic are given as below:

Location	O	E	(O - E)^2/E
North Side	70	60	1.666666667
Pleasant Township	75	60	3.75
Southwyck	70	60	1.666666667
I-90	50	60	1.666666667
Venice Avenue	35	60	10.41666667
Total	300	300	19.16666667

Test Statistic = Chi square = ∑[(O – E)^2/E] = 19.16666667

χ² = 19.17

P-value = 0.000728843

(By using Chi square table or excel)

P-value < α = 0.01

So, we reject the null hypothesis

Conclusion:

a. Reject the null hypothesis because the computed p-value is less than the level of significance.