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Business Statistics Topic 7 Chi-Squared

Business Statistics

7. Chi-Squared Goodness of Fit Tests

7.1 Introduction

The data set for a Chi-squared test is always a set of frequencies or counts. These are the observed frequencies denoted by o. The null hypothesis is a statement which, if it were true, would lead to a different set of frequencies, the expected frequencies denoted by e. The test statistic used in the chi-squared test is based on the differences between the corresponding o & e values. If the hypothesis is good all the differences will be small.

The test statistic is:

Each difference is squared and expressed as a proportion of e. If all the observed values are close to the expected values the test statistic will be small. If the hypothesis is poor (at least one large difference) the test statistic will be large. Hence for our purposes this is always a one-tailed test.

Example 1 (Worked)

The number of defective items found each day in production over a one week period, were as follows:

Day:	Mon	Tues	Wed	Thurs	Fri
Defectives:	15	8	5	5	12

Test at the 5% level of significance the null hypothesis that the differences in the number of defectives on different days are due to chance.

H0: There IS NO significant difference in the number of defectives on each day.

H1: There is a significant difference in the number of defectives on each day.

Total number of defectives in week = 45, therefore 9 defectives expected each day.

Day:	Mon	Tues	Wed	Thurs	Fri
Observed, o	15	8	5	5	12
Expected, e	9	9	9	9	9
o – e	6	-1	-4	-4	3
(o – e)2	36	1	16	16	9
(o – e)2/e	4	0.1111	1.7778	1.7778	1

= 8.6667

Degrees of Freedom -> the number of independent expected values.

Degrees of Freedom, = number of e values – number of constraints on the e values.

Here (and in many examples) the only constraint is that the expected values must add up to 45 and if 4 are known then the 5th can be found by subtracting the sum of the other 4 from 45. So Degrees of Freedom, = 4.

Critical value (from χ2 table using 0.05 column) with 4 degrees of freedom =

Test Statistic = 8.6667 < Critical Value = 9.488

So: Do NOT Reject H0 at 5% level.

SPSS

You will need to put the number of Defectives in one column, Day in another using 1=Monday etc.

Use D ata- W eight Cases, weight cases by Defectives then

Use A nalyse – N onparametric tests – L egacy Dialogs- C hi square

Enter Day into Test Variable List the click OK.

Excel

The CHITEST function takes the Observed and Expected values as arguments and returns a p-value. The Expected values must first be calculated using appropriate formulae.

Alternatively the test statistic can be calculated as in the table above.

The Critical Value can be obtained using =CHISQ.INV.RT(α, ).

In the above example α,=0.05 and = 4.

The p-value can be obtained using =CHISQ.DIST.RT(Test Statistic, ).

Note: CHIINV and CHIDIST also work as above.

Example 2

A die is thrown 300 times.

H0: The die is fair

H1: The die is biased

Score	1	2	3	4	5	6
o	40	63	48	53	38	58
e	50	50	50	50	50	50
o – e	-10	13	-2	3	-12	8
(o – e)2	100	169	4	9	144	64
(o – e)2/e	2	3.38	0.08	0.18	2.88	1.28

= 9.8

Degrees of freedom, = 6 – 1 = 5

Critical value = 11.070

so H0.

i.e. it is

7.2 Tests for Distributions

Example 3 (Worked)

The numbers of serious accidents each day on a section of motorway were recorded over a 250 day period:

No. serious accidents	0	1	2	3	4	5
No. days	100	70	45	20	10	5

Test at the 1% significance level the null hypothesis that the number of serious accidents follows a Poisson distribution.

The Poisson distribution has the following equation:

Where

k is the number of events occurring in the unit of time under consideration, in this case number of road accidents in one day. (k is a whole number)

λ is the mean value of k.

e is the constant 2.7182818….

H0: The number of serious accidents follows a Poisson distribution.

H1: The number of serious accidents does NOT follow a Poisson distribution.

Mean number of serious accidents per day, M, is (70+90+60+40+25)/250=285/250=1.14

We use M as our estimate for λ = 1.14

Using Poisson Distribution with mean 1.14:

P(0) = = 0.319819

P(1) = = 0.364594

P(2) = = 0.207818

P(3) = = 0.078971

P(4) = = 0.022507

P(5 or more) = 1-[P(0)+P(1)+ P(2)+P(3)+ P(4)]=0.00629117

Expected values are given by P(k) times number of days (250):

No. serious accidents	0	1	2	3	4	5 or more
No. days	79.955	91.148	51.955	19.743	5.627	1.573

But:

The assumptions underlying the Chi-squared test are violated if any of the expected values are substantially less than 5.

So, the “5 or more” class must be merged with the “4” class:

No. serious accidents	0	1	2	3	4 or more
No. days, o	100	70	45	20	15
Expected, e	79.955	91.148	51.955	19.743	7.199
o – e	20.045	-21.148	-6.955	0.257	7.801
(o – e)2	401.8118	447.2557	48.3665	0.0662	60.8482
(o – e)2/e	5.02549	4.90689	0.93094	0.00335	8.45175

Two constraints:

1. Must add up to 250 (days)

2. Mean = 1.14

So degrees of freedom, = 5 – 2 = 3

Critical value is from table.

Test statistic = 19.318 > 11.345 = Critical value

Reject H0 at 1% level. Also we reject at levels down to 0.1%

There is strong evidence to suggest the number of serious accidents does NOT follow a Poisson distribution.

Example 4

A random sample of 100 managers were asked to rate the success of an advertising campaign on a scale from 0 to 100.

Score	Number of managers (f)	mid-point (m)	fm	fm2
0 < 40	8	20	160	3200
40 < 50	16	45	720	32400
50 < 60	13	55	715	39325
60 < 70	23	65	1495	97175
70 < 80	15	75	1125	84375
80 < 90	16	85	1360	115600
90 – 100	9	95	855	81225
Totals	100		6430	453300

Mean = 64.3

Standard deviation = 20.063

Test at 5% level of significance the hypothesis that the scores are normally distributed.

H0: Scores are normally distributed

H1: Scores are NOT normally distributed

1. Calculate by how many standard deviations each class boundary differs from the mean (this gives us z values for the class boundaries).

2. For each class use the z table to determine the probability of an observation being in that class. Unlike the Poisson distribution there is no convenient formula.

3. From the probabilities calculate expected values for each class.

4. Using observed and expected values, calculate chi-squared.

Note: all expected values > 5 so proceed to table:

Score	< 40	40 < 50	50 < 60	60 < 70	70 < 80	80 < 90	90+
o	8	16	13	23	15	16	9
e
o – e
(o – e)2
(o – e)2/e

=

Three constraints:

1. Expected values must add up to 100

2. The mean must be 64.3

3. The standard deviation must be 20.063

So there are 7 – 3 = 4 degrees of freedom (= ).

Critical value is

so H0. It is

7.3 Contingency Tables

A contingency table is a particular kind of chi-squared test where data items are classified according to two criteria, and the null hypothesis is that the two criteria are independent of each other.

Example 5

The following table shows the absence records of 100 employees in a company for one year classified according to method of payment.

		Days Absent
		0	1 to 3	4 or more	TOTAL
Method of payment	Hourly	6	12	22
	Weekly	10	11	9
	Monthly	15	8	7
	TOTAL

Use a 5% level significance test to decide whether there is evidence to indicate a relationship between absences and method of payment.

Expected Value = Column Total X Row Total

Grand Total

Expected Values		Days Absent
		0	1 to 3	4 or more	TOTAL
Method of payment	Hourly
	Weekly
	Monthly
	TOTAL

o	e	o – e	(o – e)2	(o – e)2/e
6
10
15
12
11
8
22
9
7
			Total:

For contingency tables, degrees of freedom = (r – 1) X (c – 1)

Where: r = number of rows and c = number of columns.

SPSS

You will need to put the Frequencies (number of employees) in one column, Method of payment in another column and category of Days absent in another.

Use D ata- W eight Cases, weight cases by Frequency then

Use A nalyse - D e scriptive Statistics – C rosstabs.

Enter the appropriate fields into Row(s), Column(s) and ‘Layer 1’.

Click Statistics

Tick the box for C h i-square

Click Continue

Click OK

Example 6

In a survey designed to test the preferences for different daily tabloid newspapers 200 men and 150 women who said they read a daily tabloid newspaper regularly gave their first preferences as follows:

	Men	Women	TOTAL
Daily Star	46	37
Daily Mail	30	26
Sun	55	44
Daily Mirror	51	37
Daily Express	18	6
TOTAL

Test at the 5% level the null hypothesis that the newspaper read is independent of gender.

Expected Values	Men	Women	TOTAL
Daily Star
Daily Mail
Sun
Daily Mirror
Daily Express
TOTAL

o	e	o – e	(o – e)2	(o – e)2/e
46
30
55
51
18
37
26
44
37
6
			Total:

Example 7

The table below shows the incidence of attack by cholera on 400 inoculated people and on 300 who had not been inoculated. Use a 5% significance level test to decide whether inoculation affects proneness to attack.

	Inoculated	Not inoculated	TOTAL
Attacked	5	12
Not attacked	395	288
TOTAL

Note: As this is a 2 X 2 contingency table, use Yates’ correction:

The test statistic is now

Expected Values	Inoculated	Not inoculated	TOTAL
Attacked
Not attacked
TOTAL

o	e	|o – e|
			Total:

Example 8

In an auditing exercise 150 invoices were examined for errors. 90 of the invoices were for amounts of £50 or less and the remainder for amounts greater than £50.

	No errors	At least 1 error	TOTAL
£50 or less	70	20
Greater than £50	38	22
TOTAL

Carry out a chi-squared test at the 5% level of significance to decide whether the presence of errors is related to invoice value.

Expected Values	No errors	At least 1 error	TOTAL
£50 or less
Greater than £50
TOTAL

o	e	|o – e|
			Total:

Exercises

1. Four different makes of machine tool were tried over a period and the breakages were 24, 18, 19, 19 respectively. Is this result consistent, at the 5% level, with the hypothesis that there is no real difference between the different makes of tool?

2. An advertiser receives 60, 112 and 68 responses respectively to the same advertisement in three different papers. Is one paper significantly better than the others as an advertising medium?

3. An engineering research laboratory buys one packet of 12 bolts from each of 100 different manufacturers. Each bolt is tested to destruction and the following distribution of failures found.

Number of failures: 0 1 2 3 4 5

Frequency: 80 12 3 3 1 1

Calculate the mean number of defectives per packet and show that if all the bolts are assumed equally likely to fail the test then the probability of failure is 0.03. After suitable regrouping of the data, test at the 5% level whether the observations agree with the appropriate binomial distribution. Comment on the result.

Equation for binomial distribution is:

Where

and

4. As part of an investigation during World War II south-east London was divided up into 576 squares of side 1 km and the number of bombs falling in each such area during a certain specified time counted. It was required to test the hypothesis that bombs were being dropped randomly (i.e according to a Poisson distribution).

Bombs:	0	1	2	3	4	5 or more
Frequency:	229	211	93	35	8	0

Use a chi-squared test to decide whether the hypothesis of random bombing can be rejected at the 5% level of significance.

5. It is required to test whether there is an association between the income of salespersons of hotel supplies and their educational background. A random sample of 400 salespersons is chosen and the following results obtained:

	No university education	University but no degree	Graduate
< £10,000	63	78	29
£10,000 < £25,000	48	82	45
> £25,000	9	20	26

Test the hypothesis of no association at the 5% level.

6. Five 24-can boxes of tinned salmon were inspected for quality. The following results were obtained:

	Brand 1	Brand 2	Brand 3	Brand 4	Brand 5
Below specification	4	10	6	2	8
Adequate	20	14	18	22	16

Use a Chi-squared contingency table to test at the 5% level of significance whether the brands can be believed to be of comparable quality.

7. A large department store operating several branches wants to determine whether there is an association between an employee’s length of service and his/her opinion of the store’s annual “back to school” sale. The table below shows the judgement of a random sample of 440 employees:

	2 years or less	3 – 5 years	6 – 10 years	11 or more years
Sale very good	48	51	43	35
Sale satisfactory	22	42	59	57
Sale poor	15	22	33	13

Test the hypothesis of no association at the 5% significance level.

8. In a ‘flu epidemic, 927 children contracted the virus. 408 of them received no treatment and of these 104 suffered after-effects. Of the remainder who did receive treatment, 166 suffered after-effects. Test at the 5% significance level the hypothesis that the after-effects are unrelated to whether treatment was received and comment on the result.

9. 300 components of a particular kind were sampled at random and the dimension, X, was measured in each case:

Measurement, X mm	Number of components
0 < 2	7
2 < 4	25
4 < 6	54
6 < 8	80
8 < 10	70
10 < 12	50
12 < 14	10
14 < 16	4

a) Show that the components have mean dimension of 7.6 mm with standard deviation 2.865 mm.

b) If the dimensions were normally distributed with mean 7.6 mm and standard deviation 2.865 mm, find how many of the 300 components could be expected to have measurements in each of the 8 classes specified above.

c) Carry out a chi-squared goodness of fit test on the null hypothesis that the measurements can be believed to come from a population of normally distributed measurements.

10. A British newspaper claimed that a person’s prospects of becoming very wealthy could be influenced by the time of year they were born.

To support this claim a breakdown was published of the dates of birth of the 1,000 richest people in England. These were shown in twelve categories ‘signs of the zodiac’ used in astrology.

Investigate the validity of the newspaper’s claim.

From	To	Number Born
21 Mar	20 Apr	94
21 Apr	21 May	89
22 May	21 Jun	99
22 Jun	22 Jul	80
23 Jul	22 Aug	86
23 Aug	23 Sep	75
24 Sep	23 Oct	78
24 Oct	22 Nov	65
23 Nov	21 Dec	82
22 Dec	20 Jan	96
21 Jan	19 Feb	77
20 Feb	20 Mar	79
TOTAL:		1000

SOURCE: Daily Mail, 8th April 2009. Available on-line at:

http://www.dailymail.co.uk/news/article-2305535/Sir-Paul-McCartney-Jamie-Oliver-Richest-people-England-study-reveals-high-concentration-Geminis.html

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