Question

The following table represents the numbers of employees, by age group, who were promoted or not promoted in a sample of restaurants in last year. Using a 0.01 significance level, can you conclude

that the people in some age groups are more likely to be promoted than those in other age groups?

Age
Under 30 30-39 40-49 50 and Over
Promoted 89.0 77.0 102.0 63.0
Not Promoted 99.0 110.0 65.0 70.0

A) Calculate the table of expected values? Circle the correct table.
99.19 100.7 81.89 65.22
99.81 95.3 87.11 67.78

86.19 91.7 90.89 65.22
95.81 87.3 85.11 67.78

92.19 91.7 81.89 65.22
95.81 95.3 85.11 67.78

92.19 84.7 77.89 65.22
90.81 95.3 95.11 67.78

92.19 86.7 81.89 65.22
95.81 91.3 85.11 66.78

B) Calculate the χ2 Test Statistic:

(a) 14.82 (b) 14.768 (c) 14.68 (d) 14.658 (e) 14.542

C) Identify critical value:

(a) 12.838 (b) 9.348 (c) 7.815 (d) 6.251 (e) 11.345

D) What determination about the Claim can you make?
(a) There appears to be enough evidence to claim the the two variables are independent.
(b) There does not appear to be enough evidence to claim the the two variables are independent.
(c) We need a larger sample before we can make a determination about independence.
(d) We first need more samples since the mean of sample statistics is uniformly distributed.
(e) We do not know how accurate the sample is, so we can make no determination about the claim.

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: The people in all age groups are equally likely to be promoted among all age groups.

Alternative hypothesis: H_a: The people in some age groups are more likely to be promoted than those in other age groups.

We are given level of significance = α = 0.01

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 4

Degrees of freedom = df = (r – 1)*(c – 1) = 1*3 = 3

α = 0.01

Critical value = 11.345

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Age group
Promotion	Under 30	30 to 39	40 to 49	50 and over	Total
Promoted	89	77	102	63	331
Not promoted	99	110	65	70	344
Total	188	187	167	133	675

Expected Frequencies
	Age group
Promotion	Under 30	30 to 39	40 to 49	50 and over	Total
Promoted	92.19	91.70	81.89	65.22	331
Not promoted	95.81	95.30	85.11	67.78	344
Total	188	187	167	133	675

Calculations
(O - E)
-3.18963	-14.6993	20.10815	-2.21926
3.18963	14.69926	-20.1081	2.219259

(O - E)^2/E
0.110357	2.35627	4.937459	0.075516
0.106186	2.267225	4.750869	0.072662

Chi square = ∑[(O – E)^2/E] = 14.676544

Chi square test statistic = 14.68

P-value = 0.002115

(By using Chi square table or excel)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the people in some age groups are more likely to be promoted than those in other age groups.

A) Calculate the table of expected values?

92.19	91.70	81.89	65.22
95.81	95.30	85.11	67.78

B) Calculate the χ2 Test Statistic:

Answer: (c) 14.68

C) Identify critical value:

Answer: (e) 11.345

D) What determination about the Claim can you make?

Answer: (b) There does not appear to be enough evidence to claim the the two variables are independent.