Question

A steel company is considering the relocation of one of its manufacturing plants. The company’s executives have selected four areas that they believe are suitable locations. However, they want to determine if the average wages are significantly different in any of the locations, since this could have a major impact on the cost of production. A survey of hourly wages of similar workers in each of the four areas I performed with the following results.

	Hourly Wages ($)
	Area 1	Area 2	Area 3	Area 4
1	11	15	13	20
2	12	16	14	16
3	11	18	15	18
4	13	17	15	17
5	10	14	12	16

a. Do the data indicate a significant difference among the average hourly wages in the four areas? Construct the 10 steps of hypothesis testing using α = 0.05 to answer the question.

b. What assumptions were mad in performing the test in part a? Do the data appear to satisfy these assumptions? Explain.

Answer 1

I am using Excel to solve Part a.

a.)

The Hypotheses are:

Null: Average hourly Wages are same in all 4 locations

Ho: ₁ = ₂ = ₃ = ₄

Alternate: Average hourly Wages are different in 4 locations

Ho: ₁₂₃₄

To test the mean of these groups are statistically same or not, we will perform 1-Way Anova and check the value of P-value of F test

Given data:

-> Now Go To Data tab, Select data Analysis

-> Select one way Anova.

-> Select Group by columns and select the check box indicating that first row are labels

You will get below results

The average of means of 4 groups, Am = (11.4 + 16 + 13.8 + 17.4) / 4 = 14.65

Now, Sum of Squares between groups, SSB = n[ (Average(A1) - Am)² + (Average(A2) - Am)² + (Average(A3) - Am)² + (Average(A4) - Am)²]

SSB = 5 * [ (11.4 - 14.65)² + (16 - 14.65)² + (13.8 - 14.65)² + (17.4 - 14.65)²] = 103.35

df (Between degrees of freedom) = No of Groups - 1 = 4 - 1 = 3

MSB = SSB / df (Between degrees of freedom) = 103.35 / 3 = 34.45

SSW = [ (11.4 - 11)² + (11.4 - 12)² + (11.4 - 11)² + (11.4 - 13)² + (11.4 - 10)²] +

[ (16 - 15)² + (16 - 16)² + (16 - 18)² + (16 - 17)² + (16 - 14)²] + ..... for remaining 2 groups = 33.2

df ( Within Groups) = No of groups * ( No of terms in each groups - 1) = 4 * (5 - 1) = 16

MSW = SSW / df ( Within Groups)) = 33.2 / 16 = 2.075

F = MSB / MSW = 34.45 / 2.075 = 16.60241

F-critical (3, 16) = 3.238

Checking the F distribution table for 16.60241 we get p-value = 0.0036

Given that alpha = 0.05

Since from our test we find that, p < 0.05

So, we can reject the Null Hypotheses. Hence the Average hourly wages of 4 Areas are significantly different.

b.)

Assumptions of One Way ANOVA:

If any of the conditions are not satisfied, the results from the use of ANOVA techniques may be unreliable. The assumptions are:

1. Each sample is an independent random sample

-> There is no particular way to test this. But in our case this condition is satisfied since there are 4 different area locations

2. The population variances are equal across responses for the group levels. This can be evaluated by using the following rule of thumb: if the largest sample variance divided by the smallest sample variance is not greater than two, then assume that the population variances are equal.

In our case, 2.8 / 1.3 = 2.15 = 2(approximately)

So, this rule is also satisfied.

3.) The distribution of the response variable follows a normal distribution

We talk about the one-way ANOVA only requiring approximately normal data because it is quite "robust" to violations of normality, meaning that assumption can be a little violated and still provide valid results

The Shapiro-Wilk test tests the null hypothesis that the samples come from a normal distribution against the alternative hypothesis that the samples do not come from a normal distribution

I performed the above test for all 4 Area locations and all 4 follows normality test. Data are approximately normal. Any of the online avaialable calculators can be used to perform the test