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CHI-SQUARE ( Χ2) TESTS
DR NJOGU
11/22/2023
1
CHI-SQUARE ( Χ2) TESTS
• Chi Square test is a statistical procedure used to examine differences between
categorical data/variables in the same population.
• It follows a chi-square probability distribution.
• Tests whether a frequency distribution obtained experimentally fits an
“expected" frequency distribution"
• Expected Frequency is based;
• Theoretical or
• Previously known probability of each outcome.
11/22/2023
2
• 𝒙𝟐 =∑
(𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚)𝟐
𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
• 𝒙𝟐 =∑
(𝑶−𝑬)𝟐
𝑬
• The calculated value is compared with the tabulated value
11/22/2023
3
• Steps in Chi square
• An experiment is conducted; a simple random sample is taken from a
population, and each member of the population is grouped into one of k
categories.
• Step 1: The observed frequencies are calculated for the sample.
• Step 2: The expected frequencies are obtained from previous knowledge (or
belief) or probability theory.
• 5 products – expected 20% will go for each product
11/22/2023
4
• Step 3: a null hypothesis is adopted and Chi Square test is performed:
i. The null hypothesis H0: the population frequencies are equal to the
expected frequencies (O = E).
ii. The alternative hypothesis, Ha: the null hypothesis is false
iii. The level of significance is selected eg 95%
iv. The degrees of freedom: df = k -1. (k is number of groups)
11/22/2023
5
v. Calculate 𝒙𝟐:
𝒙𝟐 =∑
(𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐
𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅
𝒙𝟐 =∑
(𝑶−𝑬)𝟐
𝑬
(vi) From α and k - 1, a 𝒙𝟐critical value is determined from the chi-square table.
(vii) Reject H0 if χ2
calculated >𝒙𝟐
critical value
11/22/2023
6

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CHI-SQUARE ( χ2) TESTS.pptx

  • 1. CHI-SQUARE ( Χ2) TESTS DR NJOGU 11/22/2023 1
  • 2. CHI-SQUARE ( Χ2) TESTS • Chi Square test is a statistical procedure used to examine differences between categorical data/variables in the same population. • It follows a chi-square probability distribution. • Tests whether a frequency distribution obtained experimentally fits an “expected" frequency distribution" • Expected Frequency is based; • Theoretical or • Previously known probability of each outcome. 11/22/2023 2
  • 3. • 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 • 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 • The calculated value is compared with the tabulated value 11/22/2023 3
  • 4. • Steps in Chi square • An experiment is conducted; a simple random sample is taken from a population, and each member of the population is grouped into one of k categories. • Step 1: The observed frequencies are calculated for the sample. • Step 2: The expected frequencies are obtained from previous knowledge (or belief) or probability theory. • 5 products – expected 20% will go for each product 11/22/2023 4
  • 5. • Step 3: a null hypothesis is adopted and Chi Square test is performed: i. The null hypothesis H0: the population frequencies are equal to the expected frequencies (O = E). ii. The alternative hypothesis, Ha: the null hypothesis is false iii. The level of significance is selected eg 95% iv. The degrees of freedom: df = k -1. (k is number of groups) 11/22/2023 5
  • 6. v. Calculate 𝒙𝟐: 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 (vi) From α and k - 1, a 𝒙𝟐critical value is determined from the chi-square table. (vii) Reject H0 if χ2 calculated >𝒙𝟐 critical value 11/22/2023 6
  • 7. EXAMPLE A survey is carried out among 1600 coffee drinkers on how much coffee they drink in order to confirm previous studies. Previous studies indicate that 72% of Kenyans drink coffee. The results of previous studies and those of the survey are presented below. At α = 0.05, is there enough evidence to conclude that the distributions are the same? 11/22/2023 7
  • 8. Responses (O) % of coffee drinkers 2 cups per week 15% 1 cup per week 13% 1 cup per day 27% >2cups per day 45% Responses Number of coffee drinkers (E) 2 cups per week 206 1 cup per week 193 1 cup per day 462 >2cups per day 739 11/22/2023 8
  • 9. (i) The null hypothesis H0: the population frequencies (O) are equal to expected frequencies (E) (ii)The alternate hypothesis, Ha: the null hypothesis is false. (iii) α = 0.05. (iv)The degrees of freedom: df = k - 1 = 4 - 1 = 3. (v)The 𝒙𝟐 can be calculated using a table: 11/22/2023 9
  • 10. Response % of Coffee Drinkers E O O - E (O - E)2 [(O-E)2]/E 2 cups per week 15% 0.15 x 1600 = 240 206 -34 1156 4.817 1 cup per week 13% 0.13 x 1600= 208 193 -15 225 1.082 1 cup per day 27% 0.27 x 1600= 432 462 30 900 2.083 2+ cups per day 45% 0.45 x 1600 = 720 739 19 361 0.5014 ∑ (𝑶−𝑬)𝟐 𝑬 8:483 11/22/2023 10
  • 11. 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 = 8:483 i. From α = 0:05 and df = k - 1 = 3, the 𝒙𝟐 critical is 7.815. ii. s there enough evidence to reject H0? • Since χ2 calculated 8:483 > 𝒙𝟐 critical 7.815, we reject the null hypothesis 11/22/2023 11
  • 12. EXAMPLE A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly selected shoppers is conducted, and the results about which one of the stores shoppers prefer are below. Is there enough evidence at a significance level α = 0.05 to conclude that the proportions are the same? 11/22/2023 12
  • 13. Store A B C D E Number of Shoppers 262 234 204 190 210 11/22/2023 13
  • 14. (i) The null hypothesis H0: the population frequencies are equal to the expected frequencies (ii)The alternative hypothesis, Ha: the null hypothesis is false. (iii) α = 0:05. (iv)The degrees of freedom: df = k - 1 = 5 - 1 = 4. (v) 𝒙𝟐 can be calculated using a table: 11/22/2023 14
  • 15. Preference % of Shoppers E O O - E (O - E)2 [(O-E)2]/E A 20% 0.2 x 1100 = 220 262 42 1764 8.018 B 20% 0.2 x 1100 = 220 234 14 196 0.891 C 20% 0.2 x 1100 = 220 204 -16 256 1.163 D 20% 0.2 x 1100 = 220 190 -30 900 4.091 E 20% 0.2 x 1100 = 220 210 -10 100 0.455 ∑ (𝑶−𝑬)𝟐 𝑬 11/22/2023 15
  • 16. • 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 = 14.618 χ2= 14.618 i. From α = 0:05 and df=k - 1 = 4, the 𝒙𝟐 critical is 9.488. ii. Is there enough evidence to reject H0? iii. Since χ2 =14.618 > 9.488, we reject the null hypothesis 11/22/2023 16
  • 17. INDEPENDENCE • Two events are independent if the occurrence of one of the events has no effect on the occurrence of the other event. • A chi-square independence test is used to test whether or not two variables are independent. • Experiments are conducted in which the frequencies for two variables are determined. • To use the test, the same assumptions must be satisfied: the observed frequencies are obtained through a simple random sample, and each expected frequency is at least 5. • The frequencies are written down in a table: the columns contain outcomes for one variable, and the rows contain outcomes for the other variable. 11/22/2023 17
  • 18. • The procedure for the hypothesis test is essentially the same. The differences are that: i. H0 is that the two variables are independent. ii. Ha is that the two variables are not independent (they are dependent). iii. The expected frequency Er,c for the entry in row r, column c is calculated using: 𝐄𝐫, 𝐜 =∑ (𝒔𝒖𝒎 𝒐𝒇 𝒓𝒐𝒘𝒔,𝒓))(𝒔𝒖𝒎 𝒐𝒇 𝒄𝒐𝒍𝒖𝒎𝒏𝒔,𝒄) 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒊𝒛𝒆 = 8.843 i. The degrees of freedom: (no. of rows-1)x(no. of columns-1). 11/22/2023 18
  • 19. EXAMPLE The results of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table below. At α = 0.10, is there enough evidence to conclude that the treatment and result are independent? 11/22/2023 19
  • 20. Acetaminophen (c.1) Ibuprofen (c.2) Codeine (c.3) Total (r.1) Significant Improvement 58 81 61 200 (r.2) Slight Improvement 42 19 39 100 Total 100 100 100 300 11/22/2023 20
  • 21. SOLUTION • First, calculate the column and row totals. • Then find the expected frequency for each item and write it in the parenthesis next to the observed frequency. • Now perform the hypothesis test. i. The null hypothesis H0: the treatment and response are independent. ii. The alternative hypothesis, Ha: the treatment and response are dependent. iii. α = 0.10. iv. The degrees of freedom: df = (no. of rows - 1)x(no. of columns - 1) = (2 x 1) -(3 x 1) = 1 x 2 = 2. v. The 𝒙𝟐 critical can be calculated using a table: 11/22/2023 21
  • 22. Row, Column E O O - E (O - E)2 [(O-E)2]/E 1,1 [200x100]/300 = 66.7 58 -8.7 75.69 1.135 1,2 [200x100]/300 = 66.7 81 14.3 204.49 3.067 1,3 [200x100]/300 = 66.7 61 -5.7 32.49 0.487 2,1 [100x100]/300 = 33.3 42 8.7 75.69 2.271 2,2 [100x100]/300 = 33.3 19 -14.3 204.49 6.135 2,3 [100x100]/300 = 33.3 39 5.7 32.49 0.975 11/22/2023 22
  • 23. 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 𝛘𝟐 = 14.07: • • (vi) From α = 0:10 and d.f = 2, the critical value is 4.605. • (vii) Is there enough evidence to reject H0? Since 𝛘𝟐 14:07 > 4:605, there is enough statistical evidence to reject the null hypothesis. 11/22/2023 23
  • 24. EXAMPLE The side effects of a new drug are being tested against a placebo. A simple random sample of 565 patients yields the results below. At a significance level of α = 0:05, is there enough evidence to conclude that the treatment is independent of the side effect of nausea? • 11/22/2023 24
  • 25. Result Drug (c.1) Placebo (c.2) Total Nausea (r.1) 36 13 49 No nausea (r.2) 254 262 516 Total 290 275 565 11/22/2023 25
  • 26. i. The null hypothesis H0: the treatment and response are independent. ii. The alternative hypothesis, Ha: the treatment and response are dependent. iii. α = 0:01. iv. The degrees of freedom: (no. of rows-1)x(no. of columns-1) = (2 - 1) x (2 - 1) = 1 x 1 = 1. v. The test statistic can be calculated using a table: 11/22/2023 26
  • 27. Row, Column E O O – E (O - E)2 [(O-)2]/E 1,1 [49x290]/565 = 25.15 36 10.85 117.72 4.681 1,2 [49x275]/565 = 23.85 13 -10.85 117.72 4.936 2,1 [516x290]/565 = 264.85 254 -10.85 117.72 0.444 2,2 [516x275]/565 = 251.15 262 10.85 117.72 0.469 11/22/2023 27
  • 28. 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 𝛘𝟐 = 10.53: • i. From α = 0:10 and d.f = 1, the critical value is 2.706. ii. Is there enough evidence to reject H0? Since 𝛘𝟐 10.53 > 2.706, there is enough statistical evidence to reject the null hypothesis and to believe that there is a relationship between the treatment and response. 11/22/2023 28
  • 29. EXAMPLE • Suppose that we have a 6-sided die. We assume that the die is unbiased (upon rolling the die, each outcome is equally likely). An experiment is conducted in which the die is rolled 240 times. The outcomes are in the table below. At a significance level of α = 0.05, is there enough evidence to support the hypothesis that the die is unbiased? 11/22/2023 29
  • 30. Outcome 1 2 3 4 5 6 Frequency 34 44 30 46 51 35 11/22/2023 30
  • 31. i. The null hypothesis H0: each face is equally likely to be the outcome of a single roll. ii. The alternative hypothesis, Ha: the null hypothesis is false. iii. α = 0:05. iv. The degrees of freedom: k - 1 = 6 - 1 = 5. v. The test statistic can be calculated using a table: 11/22/2023 31
  • 32. Face E O O – E (O - E)2 (O-E)2/E 1 240/6 = 40 34 -6 36 0.9 2 240/6 = 40 44 4 16 0.4 3 240/6 = 40 30 -10 100 2.5 4 240/6 = 40 46 6 36 0.9 5 240/6 = 40 51 11 121 3.025 6 240/6 = 40 35 -5 25 0.625 11/22/2023 32
  • 33. 𝒙𝟐 =∑ (𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 −𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅)𝟐 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒙𝟐 =∑ (𝑶−𝑬)𝟐 𝑬 𝛘𝟐 = 15.086: i. From α= 0:01 and k - 1 = 6, the critical value is 15.086. ii. Is there enough evidence to reject H0? Since 𝛘𝟐 8:35 < 15:086, we fail to reject the null hypothesis, that the die is fair. 11/22/2023 33