14 Chi-squared Test
14.1 Goodness of Fit Test
In this section we use a chi-square test to determine whether a population being sampled has a specific probability distribution.
14.1.1 A Multinomial Population
Multinomial Experiment
A multinomial experiment is one that possesses the following properties.
The experiment consists of a fixed number \(n\) of trials.
The outcome of each trial can be classified into one of \(k\) categories, called \(cells\).
The probability \(p_i\) that the outcome will fall into cell \(i\) remains constant for each trial. Moreover, \(p_1 + p_2 + ...+ p_k = 1\)
Each trial of the experiment is independent of the other trials.
Testing Market Shares
Company A has recently conducted aggressive advertising campaigns to maintain and possibly increase its share of the market (currently \(45\%\)) for fabric softener. Its main competitor, company B, has \(40\%\) of the market, and a number of other competitors account for the remaining \(15\%\).
To determine whether the market shares changed after the advertising campaign, the marketing manager for company A solicited the preferences of a random sample of 200 customers of fabric softener.
Of the 200 customers, 102 indicated a preference for company A’s product, 82 preferred company B’s fabric softener, and the remaining 16 preferred the products of one of the competitors. Can the analyst infer at the \(5\%\) significance level that customer preferences have changed from their levels before the advertising campaigns were launched?
We recognize this experiment as a multinomial experiment, and we identify the technique as the chi-squared goodness-of-fit test. Because we want to know whether the market shares have changed, we specify those precampaign market shares in the null hypothesis.
\[ H_0: p_1=0.45;\ \ p_2=0.40; \ \ p_3=0.15 \]
The alternative hypothesis attempts to answer our question, Have the proportions changed? Thus,
\[ H_1: At \ \ least \ \ one \ \ p_i \ \ is \ \ not\ \ equal\ \ to\ \ its\ \ specified\ \ value \]
Chi-Squared Goodness-of-Fit Test Statistic
\[ \chi^2 =\sum_{i=1}^k \frac{(f_i-e_i)^2}{e_i} \] \(Where, \ \ f_i=observed \ \ frequency \ \ and \ \ e_i=expected \ \ frequency\)
Note that, \(e_i=n*p_i\)
The sampling distribution of the test statistic is approximately chi-squared distributed with \(k-1\) degrees of freedom, provided that the sample size is large.
Test Statistic calculation
| Company | Observed frequency, \(f_i\) | Expected frequency, \(e_i\) | \((f_i-e_i)\) | \(\frac{(f_i-e_i)^2}{e_i}\) |
|---|---|---|---|---|
| A | 102 | 90 | 12 | 1.60 |
| B | 82 | 80 | 2 | 0.05 |
| Other | 16 | 30 | -14 | 6.53 |
| Total | 200 | 200 | \(\chi^2=8.18\) |
Critical value
At \(\alpha =0.05\) and for \(df=3-1=2\), \(\chi^2_\alpha=5.99\) .
Decision
Since \(\chi^2 > \chi^2_{\alpha}\) so reject null hypothesis.
Interpretation/ Conclusion
There is sufficient evidence at the \(5\%\) significance level to infer that the proportions have changed since the advertising campaigns were implemented.
14.1.2 Normal population (continuous)
To test whether a variable follows normal distribution with mean \(\mu\) and variance \(\sigma^2\) we will illustrate the following example.
Example A random sample of 500 car batteries was taken and the life of each battery was measured. Letting X denote battery life in years, suppose that the sample revealed the following distribution of battery life:
| Life (in years) | Frequency |
|---|---|
| \(X<1\) | 12 |
| \(1<X \le 2\) | 94 |
| \(2<X \le 3\) | 170 |
| \(3 <X \le 4\) | 188 |
| \(4<X \le 5\) | 28 |
| \(5<X\) | 8 |
| 500 |
Based on this data, test whether battery life follows a normal distribution with \(\mu = 2.8\) and \(\sigma^2 = 1.1^2\). Clearly state your hypotheses and use a significance level of \(\alpha = 5\%\).
Solution:
Hypotheses
H0: The battery life follows a normal distribution with \(\mu = 2.8\) and \(\sigma^2 = 1.1^2\)..
Ha:The battery life does not follow a normal distribution.
Test Statistic calculation
| Life (in years) | Probability | \(e_i=np_i\) | \(f_i\) | \(\frac{(f_i-e_i)^2}{e_i}\) |
|---|---|---|---|---|
| \(X<1\) | \(P(X<1)\) \(=P(Z<-1.64)\) \(=0.0505\) |
25.25 | 12 | 6.9530 |
| \(1<X \le 2\) | \(P(1<X \le 2)\) \(=P(-1.64<Z\le -0.73)\) \(=0.1826\) |
91.30 | 94 | 0.0798 |
| \(2<X \le 3\) | 0.3386 | 169.30 | 170 | 0.0029 |
| \(3 <X \le 4\) | 0.2902 | 145.10 | 188 | 12.6837 |
| \(4<X \le 5\) | 0.1149 | 57.45 | 28 | 15.0966 |
| \(5<X\) | 0.0228 | 11.40 | 8 | 1.0140 |
| \(\chi^2=35.83\) |
Critical value
At \(\alpha=0.05\) , and for \(df\)=6-1=5, \(\chi^2_{\alpha,5}=11.1\)
Decision
Since \(\chi^2>\chi^2_{\alpha,5}\) so we can reject the null hypothesis.
14.1.3 Uniform distribution (continuous)
To test whether a variable follows uniform distribution between \(a\) to \(b\) we will illustrate the following example.
Example Suppose X, the amount of time a person stays in bed after their alarm goes off, is uniformly distributed between 5 and a minutes. Also, suppose Y , the number of minutes they are late to work is uniformly distributed between 0 and b minutes. Over 100 days, how long they slept past their alarm (X) and how late they were to work (Y ) were recorded (in minutes). However, only the number of days for which X and Y fell within certain ranges was reported in the table below:
| 5<X<7 | 7<X<8 | 8<X<10 | Totals | |
|---|---|---|---|---|
| 0<Y<2 | 18 | 9 | 12 | |
| 2<Y<3 | 9 | 4 | 12 | |
| 3<Y<5 | 13 | 9 | 14 | |
| Totals | 100 |
(a) Based on the data given above, test whether \(a\) is equal to 10. Clearly state your hypotheses and use a significance level of \(\alpha = 5\%\).
Solution:
If the X, the amount of time a person stays in bed after their alarm goes off, is uniformly distributed between 5 and \(a\) minutes then the data will fit the uniform distribution with parameter 5 to a=10 minutes. So following hypotheses can be formed:
Hypothesis
H0: The amount of time a person stays in bed after their alarm goes off, is uniformly distributed between 5 and \(a=10\) minutes
Ha: The amount of time a person stays in bed after their alarm goes off, is NOT uniformly distributed between 5 and \(a=10\) minutes.
Test statistic calculation
If \(X\sim U(5,10)\) then PDF, \[f(x)=\frac{1}{10-5}=\frac{1}{5};\ \ 5<x<10\]
So, \(P(5<X<7)=(7-5)*\frac{1}{5}=\frac{2}{5}\) and so on
| Bed time (in mins) | \(f_i\) | \(p_i\) | \(e_i=np_i\) | \(\frac{(f_i-e_i)^2}{e_i}\) |
|---|---|---|---|---|
| 5<X<7 | 40 | \(\frac{2}{5}\) | 40 | 0.00 |
| 7<X<8 | 22 | \(\frac{1}{5}\) | 20 | 0.20 |
| 8<X<10 | 38 | \(\frac{2}{5}\) | 40 | 0.10 |
| Totals | 100 | \(\chi^2=0.3\) |
Critical value
At \(\alpha=5\%\) and \(df=3-1=2\), \(\chi^2_{\alpha,2}=5.99\).
Decision
Since, \(\chi^2<\chi^2_{\alpha,2}\) so we cannot reject null hypothesis.
14.2 Test for Independence (Categorical Data)
Consider the following example:
In an experiment to study the dependence of hypertension on smoking habits, the following data were taken on 180 individuals:
| Non-smokers | Moderate Smokers | Heavy Smokers | |
|---|---|---|---|
| Hypertension | 21 | 36 | 30 |
| No hypertension | 48 | 26 | 19 |
Test the hypothesis that the presence or absence of hypertension is independent of smoking habits. Use a 0.05 level of significance.
Solution:
We have to test the following hypothesis:
\(H_0:The \ \ column\ \ variable\ \ is\ \ independent\ \ of\ \ the\ \ row\ \ variable\)
\(H_a:The \ \ column\ \ variable\ \ is\ \ not\ \ independent\ \ of\ \ the\ \ row\ \ variable\)
Test statistic
\[ \chi^2=\sum_{i=1}^r\sum_{j=1}^c \frac {(f_{ij}-e_{ij})^2}{e_{ij}} \]
The sampling distribution of the test statistic is approximately chi-squared distributed with \((r-1)\times (c-1)\) degrees of freedom, provided that the sample size is large.
Note
\(f_{ij}=\) Observed frequency of \((i,j)^{th}\) cell;
\(e_{ij}=\)Expected frequency of \((i,j)^{th}\) cell=\(\frac{Row \ \ i \ \ total \times Column \ \ j \ \ total}{Sample \ \ size (n)}\)
Table: Contingency table with Row total and Column total
| Non-smokers | Moderate Smokers | Heavy Smokers | Row Total | |
|---|---|---|---|---|
| Hypertension | 21 | 36 | 30 | 87 |
| No hypertension | 48 | 26 | 19 | 93 |
| Column Total | 69 | 62 | 49 | 180 |
For example,
\[ \boldsymbol {e_{11}=\frac{87\times69}{180}} \]
Chi-square Statistic calculation
| Observed, \(f_i\) | Expected, \(e_i\) | \(\frac{(f_i-e_i)^2}{e_i}\) |
|---|---|---|
| 21 | 33.35 | 4.57 |
| 36 | 29.97 | 1.21 |
| 30 | 23.68 | 1.68 |
| 48 | 35.65 | 4.28 |
| 26 | 32.03 | 1.14 |
| 19 | 25.32 | 1.14 |
| \(\chi^2=14.46\) |
Critical value
At \(\alpha =0.01\) and with \(df=(2-1)*(3-1)=2\), \(\chi^2_\alpha =9.21\)
Decision
Since \(\chi^2 > \chi^2_\alpha\) so reject the null hypothesis.
Interpretation/conclusion
There is sufficient evidence at the \(5\%\) significance level to infer that the smoking habits is not independent of the the presence or absence of hypertension, rather the two variables are associated.