{"id":1996,"date":"2023-11-26T14:45:45","date_gmt":"2023-11-26T14:45:45","guid":{"rendered":"https:\/\/myknowledgehub.org\/?p=1996"},"modified":"2024-01-05T13:51:14","modified_gmt":"2024-01-05T13:51:14","slug":"research-methodology-chapter-11-2","status":"publish","type":"post","link":"https:\/\/myknowledgehub.org\/index.php\/2023\/11\/26\/research-methodology-chapter-11-2\/","title":{"rendered":"Research Methodology Chapter 11.2"},"content":{"rendered":"\t\t
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Chi-Square Test<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t
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A\u00a0chi-square<\/i><\/b>\u00a0(\u03c72<\/sup><\/i><\/b>)\u00a0statistic<\/i><\/b>\u00a0is a test that measures how expectations compare to\u00a0actual observed data (or model results). The data used in calculating a
chi-square statistic must be\u00a0random<\/b>,\u00a0raw<\/b>,\u00a0mutually exclusive<\/b>,\u00a0drawn from independent variables<\/b>, and\u00a0drawn from a large enough\u00a0sample<\/b>.<\/p>

That is, the\u00a0chi-square\u00a0<\/i><\/b>(\u03c72<\/sup><\/i><\/b>)\u00a0tests<\/i><\/b>\u00a0are certain types of\u00a0statistical hypothesis tests that are valid to perform when the test statistic\u00a0is\u00a0chi-squared distributed<\/i><\/b>\u00a0under the\u00a0null hypothesis<\/i><\/b>.<\/p>

In the standard applications of this\u00a0test, the observations are classified into mutually exclusive classes.\u00a0If\u00a0the so-called null hypothesis is true<\/i><\/b>, the\u00a0test statistic\u00a0computed from the observations follows a \u03c72<\/sup>\u00a0distribution<\/i><\/b>.
The purpose of the test is to evaluate how likely the observed frequencies
would be assuming the null hypothesis is true. Test statistics that follow a\u00a0\u03c72<\/sup>
distribution<\/i><\/b> occurs when the observations are independent and normally
distributed, which assumptions are often justified under the central limit theorem. There are also\u00a0\u03c72<\/sup>\u00a0tests<\/i><\/b>\u00a0for testing the\u00a0null hypothesis of independence of a pair of random variables based on observations of the pairs<\/i><\/b>.<\/p>

It determines whether or not the sampling distribution (if the null hypothesis is true) of the test statistic approximates a chi-squared distribution more and more closely as sample sizes increase.<\/p>

Types of chi-square<\/i><\/b>\u00a0(\u03c72<\/sup><\/i><\/b>)\u00a0tests<\/i><\/b>:\u00a0<\/p>

There are two types of chi-square\u00a0tests. Both use the chi-square statistic and distribution for different\u00a0purposes:<\/p>

1. A\u00a0<\/span>chi-square test for independence<\/i><\/b>\u00a0compares two\u00a0<\/span>variables in a contingency table to see if they are related. In a more general\u00a0<\/span>sense, it tests to see whether distributions of categorical variables differ\u00a0<\/span>from each another.<\/span><\/p>

2. A\u00a0chi-square goodness-of-fit test<\/i><\/span>\u00a0determines if a\u00a0sample data matches a population<\/i>. This test is also referred to as\u00a0Goodness-of-Fit Test<\/i><\/span>.<\/p>

\u00a0<\/div>

A\u00a0very small<\/i><\/b>\u00a0chi square test\u00a0statistic<\/i><\/b>\u00a0means that\u00a0your observed data fits your expected data extremely well<\/i>. In other words,\u00a0there is a relationship<\/i><\/b>.<\/p>

A\u00a0very large chi-square test statistic<\/i><\/b>\u00a0means that\u00a0the data does not fit very well<\/i>. In other words, there isn\u2019t a relationship<\/i><\/b>.<\/p>

\u00a0<\/i><\/b><\/p>

Assumptions<\/i><\/b>:\u00a0<\/p>

Like so many of our inference procedures, chi-square tests too have some underlying assumptions which should be in place to make the results of calculations completely trust worthy. They include:\u00a0<\/p>

1 The data in the cells should be frequencies<\/i><\/b>,\u00a0 or counts of cases <\/i><\/b>rather than percentages or some other transformation of the data.<\/p>

2. The levels<\/i><\/b>\u00a0 (or categories<\/i><\/b>) of the variables are mutually exclusive<\/i><\/b>.\u00a0
That is,\u00a0 a\u00a0 particular subject fits into one and only one level of each of the variables.<\/p>

3.\u00a0 Each subject may contribute data to one and only one cell in the \u03c72<\/sup><\/i>. If, for\u00a0example, the same subjects are tested over time such that the comparisons are\u00a0of the same subjects at Time 1, Time 2, Time 3, etc., then\u00a0\u03c72 <\/sup><\/i>may not be used.<\/p>

4. The study groups must be independent<\/i><\/b>. This means that a different test must be used if the two groups are related.\u00a0 For\u00a0 example,\u00a0 a\u00a0
different test must be used if the researcher\u2019s data consists of paired samples, such as in studies in which a parent is paired with his or her child.<\/i><\/b><\/p>

5. There are 2 variables<\/i><\/b>, and both are measured as categories<\/i><\/b>,\u00a0usually at the\u00a0nominal level<\/i><\/b>. However, data may be ordinal\u00a0data<\/i><\/b>. Interval or ratio data that have been collapsed into ordinal categories may also be used. While Chi-square has no rule about limiting the number of cells (by limiting the number of categories for each variable), a very large number of cells (over 20) can make it difficult to meet assumption #6 below, and to interpret the meaning of the results.\u00a0<\/p>

6. The value of the cell expected should be\u00a0 5 or more in at least\u00a0 80%\u00a0 of the cells,\u00a0 and no cell should have an expected of less than one (3). This assumption is most likely to be met if the sample size equals at least the number of cells multiplied by 5. Essentially, this assumption specifies the number of cases (sample size) needed to use the \u03c72<\/sup><\/i>\u00a0for any number of cells in that\u00a0\u03c72<\/sup><\/i>.\u00a0<\/p>

\u00a0<\/i><\/b><\/p>

Hypotheses<\/i><\/b>:<\/p>

Null Hypothesis (H0<\/sub>): There is “no\u00a0change” or “no difference” in situation.<\/p>

Alternative Hypothesis (H1<\/sub>): There is a\u00a0“change” or “difference” in situation.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t

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I. Test of Independence<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t
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When considering student sex and course choice, a \u03c72 test for independence could be used. To do this test, the researcher would collect data on the two chosen variables (sex and courses picked) and then compare the frequencies at which male and female students select among the offered classes using the formula given above and a \u03c72 statistical table.<\/p>

If there is no relationship between sex and course selection (that is, if they are independent), then the actual frequencies at which male and female students select each offered course should be expected to be approximately equal, or conversely, the proportion of male and female students in any selected course should be approximately equal to the proportion of male and female students in the sample.<\/p>

A \u03c72 test for independence can tell us how likely it is that random chance can explain any observed difference between the actual frequencies in the data and these theoretical expectations.<\/p>

Problem <\/strong><\/p>

Imagine you have surveyed 200 individuals to determine if there is a significant association between gender and preference for a particular smartphone brand.<\/p>

Solution<\/strong><\/p>

Step 1<\/em><\/strong>: Formulate Hypotheses<\/p>

Null Hypothesis (H0<\/sub>): There is no association between gender and smartphone brand preference.<\/p>

Alternative Hypothesis (Ha<\/sub>): There is a significant association between gender and smartphone brand preference.<\/p>

Step 2<\/em><\/strong>: Collect Data<\/p>

\u00a0<\/strong><\/p><\/td>

iPhone<\/strong><\/p><\/td>

Samsung<\/strong><\/p><\/td>

Other<\/strong><\/p><\/td><\/tr>

Male<\/strong><\/p><\/td>

30<\/p><\/td>

40<\/p><\/td>

10<\/p><\/td><\/tr>

Female<\/strong><\/p><\/td>

20<\/p><\/td>

50<\/p><\/td>

50<\/p><\/td><\/tr><\/tbody><\/table>

Step 3<\/em><\/strong>: Set Significance Level<\/p>

Choose a significance level (commonly 0.05) to determine if the observed association is statistically significant.<\/p>

Step 4<\/em><\/strong>: Create a Contingency Table<\/p>

Sum the rows and columns and create a contingency table:<\/p>

\u00a0<\/strong><\/p><\/td>

iPhone<\/strong><\/p><\/td>

Samsung<\/strong><\/p><\/td>

Other<\/strong><\/p><\/td>

Row Total<\/strong><\/p><\/td><\/tr>

Male<\/strong><\/p><\/td>

30<\/p><\/td>

40<\/p><\/td>

10<\/p><\/td>

80<\/strong><\/p><\/td><\/tr>

Female<\/strong><\/p><\/td>

20<\/p><\/td>

50<\/p><\/td>

50<\/p><\/td>

120<\/strong><\/p><\/td><\/tr>

Column Total<\/strong><\/p><\/td>

50<\/strong><\/p><\/td>

90<\/strong><\/p><\/td>

60<\/strong><\/p><\/td>

200<\/strong><\/p><\/td><\/tr><\/tbody><\/table>

Step 5<\/em><\/strong>: Calculate Expected Frequencies<\/p>

Calculate the expected frequency for each cell using the formula:<\/p>

Expected Frequency =\u00a0\"\"<\/p>

Calculation for the cell in the first row, first column (iPhone, Male)<\/p>

Expected Frequency =\u00a0\"\"<\/p>

Expected Frequency = 20<\/p>

Repeat this calculation for each cell in the table.<\/p>

In the following Table the Expected Frequency calculated are given in parenthesis in each cell.<\/p>

\u00a0<\/strong><\/p><\/td>

iPhone<\/strong><\/p><\/td>

Samsung<\/strong><\/p><\/td>

Other<\/strong><\/p><\/td>

Row Total<\/strong><\/p><\/td><\/tr>

Male<\/strong><\/p><\/td>

30 (20)<\/strong><\/p><\/td>

40 (36)<\/strong><\/p><\/td>

10 (24)<\/strong><\/p><\/td>

80<\/strong><\/p><\/td><\/tr>

Female<\/strong><\/p><\/td>

20 (30)<\/strong><\/p><\/td>

50 (54)<\/strong><\/p><\/td>

50 (36)<\/strong><\/p><\/td>

120<\/strong><\/p><\/td><\/tr>

Column Total<\/strong><\/p><\/td>

50<\/strong><\/p><\/td>

90<\/strong><\/p><\/td>

60<\/strong><\/p><\/td>

200<\/strong><\/p><\/td><\/tr><\/tbody><\/table>

Step 6<\/em><\/strong>: Calculate Chi-Square (\u03c72<\/sup>) Statistic<\/p>

\"\"<\/p>

Where, O is the observed frequency, and E is the expected frequency.<\/p>

For the given example, you would calculate contributions for each cell and sum them to get the chi-square value. In the Table below, value in parenthesis is the contributions for each cell.<\/p>

\u00a0<\/strong><\/p><\/td>

iPhone<\/strong><\/p><\/td>

Samsung<\/strong><\/p><\/td>

Other<\/strong><\/p><\/td>

Row Total<\/strong><\/p><\/td><\/tr>

Male<\/strong><\/p><\/td>

(30 \u2013 20)2<\/sup>\/20 (5)<\/strong><\/p><\/td>

(40 \u2013 36)2<\/sup>\/36 (0.44)<\/strong><\/p><\/td>

(10 \u2013 24)2<\/sup>\/24 (8.16)<\/strong><\/p><\/td>

80<\/strong><\/p><\/td><\/tr>

Female<\/strong><\/p><\/td>

(20 \u2013 30)2<\/sup>\/30 (3.33)<\/strong><\/p><\/td>

(50 \u2013 54)2<\/sup>\/54 (0.29)<\/strong><\/p><\/td>

(50 \u2013 36)2<\/sup>\/36 (5.4)<\/strong><\/p><\/td>

120<\/strong><\/p><\/td><\/tr>

Column Total<\/strong><\/p><\/td>

50<\/strong><\/p><\/td>

90<\/strong><\/p><\/td>

60<\/strong><\/p><\/td>

200<\/strong><\/p><\/td><\/tr><\/tbody><\/table>

\u03c72<\/sup> = 5 + 0.44 + 8.16 + 3.33 + 0.29 + 5.4<\/p>

\u03c72<\/sup> = 22.62<\/p>

Step 7<\/em><\/strong>: Determine Degrees of Freedom<\/p>

Degrees of freedom (df) is calculated as df = (r\u22121) \u00d7 (c\u22121),<\/p>

Where, r is the number of rows and c is the number of columns.<\/p>

For our example,<\/p>

df = (2\u22121) \u00d7 (3\u22121) = 2.<\/p>

Step 8<\/em><\/strong>: Find Critical Value or P-value<\/p>

Using the chi-square distribution table or a statistical software, find the critical value or p-value corresponding to the degrees of freedom and chosen significance level.<\/p>

Using a chi-square distribution table or statistical software, the critical chi-square value for df=2 at a significance level of 0.05 is approximately 5.99<\/strong>.<\/p>

Step 9<\/em><\/strong>: Make a Decision<\/p>

Compare the calculated chi-square value with the critical value or use the p-value to determine whether to reject the null hypothesis.<\/p>

If the calculated chi-square value is greater than the critical value or the p-value is less than the significance level, reject the null hypothesis.<\/p>

In our example, the calculated chi-square value (22.62) is greater than the critical value (5.99), hence we reject the null hypothesis, and confirm that there is a significant association between gender and smartphone brand preference.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t

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II. Goodness-of-Fit Test <\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t
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\u03c72 provides a way to test how well a sample of data matches the (known or assumed) characteristics of the larger population that the sample is intended to represent. This is known as goodness of fit. If the sample data do not fit the expected properties of the population that we are interested in, then we would not want to use this sample to draw conclusions about the larger population.<\/p>

Problem<\/strong><\/p>

Imagine you are studying the wing colour variation in a population of 120 Monarch butterflies in a local meadow to determine if the wing colour follows the ratio 3:1:2 for orange, black, and white variations, respectively.<\/p>

Solution<\/strong><\/p>

Step 1<\/em><\/strong>: Formulate Hypotheses<\/p>

Null Hypothesis (H0<\/sub>): The proportion of Monarch butterflies with each color variant follows the 3:1:2 ratio (60%, 20%, and 40%).<\/p>

Alternative Hypothesis (Ha<\/sub>): The proportion of Monarch butterflies with each colour variant deviates significantly from the suspected ratio.<\/p>

Step 2<\/em><\/strong>: Collect Data<\/p>

Color<\/strong><\/p><\/td>

Observed Frequency<\/strong><\/p><\/td><\/tr>

Orange<\/strong><\/p><\/td>

60<\/p><\/td><\/tr>

Black<\/strong><\/p><\/td>

25<\/p><\/td><\/tr>

White<\/strong><\/p><\/td>

35<\/p><\/td><\/tr><\/tbody><\/table>

Step 3<\/em><\/strong>: Set Significance Level<\/p>

Choose a significance level (commonly 0.05) to determine if the observed association is statistically significant.<\/p>

Step 4<\/em><\/strong>: Calculate Expected Frequencies<\/p>

Based on the 3:1:2 ratio, the expected frequencies are calculated using the expected percentage for each category using the following formula.\u00a0<\/p>

Expected Frequency = expected percentage * 120<\/p>

Thus, for orange, the expected frequency will be (3\/6) * 120 = 60, for black the expected frequency will be (1\/6) * 120 = 20, and\u00a0<\/span>for black the expected frequency will be (2\/6) * 120 = 40.<\/span>\u00a0<\/span>\u00a0<\/span><\/p>

In the following Table the Expected Frequency calculated are given.<\/span><\/p>

Color<\/strong><\/p><\/td>

Observed Frequency<\/strong><\/p><\/td>

Expected Frequency<\/strong><\/p><\/td><\/tr>

Orange<\/strong><\/p><\/td>

60<\/p><\/td>

60<\/p><\/td><\/tr>

Black<\/strong><\/p><\/td>

25<\/p><\/td>

20<\/p><\/td><\/tr>

White<\/strong><\/p><\/td>

35<\/p><\/td>

40<\/p><\/td><\/tr><\/tbody><\/table>

Step 6<\/em><\/strong>: Calculate Chi-Square (\u03c72<\/sup>) Statistic<\/p>

\"\"<\/p>

Where, O is the observed frequency, and E is the expected frequency.<\/p>

For the given example, you would calculate contributions for each cell and sum them to get the chi-square value. In the Table below, value in parenthesis is the contributions for each cell.<\/p>

Color<\/strong><\/p><\/td>

Chi-Square Value<\/strong><\/p><\/td><\/tr>

Orange<\/strong><\/p><\/td>

(60 \u2013 60)2<\/sup>\/60<\/p>

(0)<\/strong><\/p><\/td><\/tr>

Black<\/strong><\/p><\/td>

(25 \u2013 20)2<\/sup>\/20<\/p>

(1.25)<\/strong><\/p><\/td><\/tr>

White<\/strong><\/p><\/td>

(35 \u2013 40)2<\/sup>\/40<\/p>

(0.625)<\/strong><\/p><\/td><\/tr><\/tbody><\/table>

\u03c72<\/sup> = 0 + 1.25 + 0.625<\/p>

\u03c72<\/sup> = 1.875<\/p>

Step 7<\/em><\/strong>: Determine Degrees of Freedom<\/p>

Degrees of freedom (df) is calculated as df = n \u2212 1,<\/p>

Where, n is the number of categories.<\/p>

For our example,<\/p>

df = 3 \u2013 1 = 2.<\/p>

Step 8<\/em><\/strong>: Find the Critical Value or P-value<\/p>

Using the chi-square distribution table or a statistical software, find the critical value or p-value corresponding to the degrees of freedom and chosen significance level.<\/p>

Using a chi-square distribution table or statistical software, the critical chi-square value for df=2 at a significance level of 0.05 is approximately 5.99 <\/strong>and the p value is approximately 0.391<\/strong>.<\/p>

Step 9<\/em><\/strong>: Make a Decision<\/p>

Compare the calculated chi-square value with the critical value or use the p-value to determine whether to reject the null hypothesis.<\/p>

If the calculated chi-square value is greater than the critical value or the p-value is less than the significance level, reject the null hypothesis.<\/p>

In our example, the calculated chi-square value (1.875) is lesser than the critical value (5.99) and the p value (0.391) is greater than the significance level, hence we fail to reject the null hypothesis and confirm that there is not enough evidence to conclude that the proportion of Monarch butterflies with each colour variant deviates significantly from the suspected 3:1:2 ratio at the 5% significance level.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t

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\n\t\t\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\t\t\tBack to the Content<\/span>\n\t\t\t\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

Chi-Square Test A\u00a0chi-square\u00a0(\u03c72)\u00a0statistic\u00a0is a test that measures how expectations compare to\u00a0actual observed data (or model results). The data used in calculating achi-square statistic must be\u00a0random,\u00a0raw,\u00a0mutually exclusive,\u00a0drawn from independent variables, and\u00a0drawn from a large enough\u00a0sample. That is, the\u00a0chi-square\u00a0(\u03c72)\u00a0tests\u00a0are certain types of\u00a0statistical hypothesis tests that are valid to perform when the test statistic\u00a0is\u00a0chi-squared distributed\u00a0under the\u00a0null hypothesis. In …<\/p>\n

Research Methodology Chapter 11.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","footnotes":""},"categories":[1],"tags":[],"class_list":["post-1996","post","type-post","status-publish","format-standard","hentry","category-blog"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/posts\/1996","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/comments?post=1996"}],"version-history":[{"count":31,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/posts\/1996\/revisions"}],"predecessor-version":[{"id":2403,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/posts\/1996\/revisions\/2403"}],"wp:attachment":[{"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/media?parent=1996"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/categories?post=1996"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/myknowledgehub.org\/index.php\/wp-json\/wp\/v2\/tags?post=1996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}