11.2: Test for Independence
- Page ID
- 58316
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Understand how to perform a chi-square test for independence.
- Evaluate whether two categorical variables are related or independent.
- Use a contingency table to compare observed and expected frequencies.
- Apply the test to analyze relationships in survey data or between traits in populations.
- Interpret the results to determine the presence or absence of an association between variables.
Use the Chi-Square Test for Independence to determine whether two categorical (qualitative) variables are associated. Qualitative data involves characteristics or categories (such as names or labels), and we analyze it by counting how many individuals fall into each category.
For example, suppose researchers are investigating whether there is a relationship between breastfeeding duration and the likelihood of being diagnosed with autism spectrum disorder (ASD). They would collect data on how long each mother breastfed her child and whether the child was later diagnosed with ASD. This information would be organized into a contingency table, showing the frequency of individuals across the combined categories.
The test then evaluates whether the distribution of counts in each table cell differs from what we would expect if the two variables were truly independent. In this context, independence means that breastfeeding does not affect whether a child develops ASD.
The goal of the test is to see if there is evidence of dependence—that is, whether one variable affects the other. This is framed statistically as:
- Null hypothesis (H₀): The variables are independent.
- Alternative hypothesis (H₁): The variables are dependent.
To test this, we use the Chi-Square Test for Independence, which compares observed counts to expected counts under the assumption of independence.
Relation to Independent Probability Rule
For each \(\chi^2\) test for independence, data is collected, counted, and then organized into a contingency table. These values are known as the observed frequencies, and the symbol for an observed frequency is \(O\). Total each row and column. The null hypothesis is that the two variables are independent. If two events are independent then \(P(B) = P(B | A)\) and we can use the multiplication rule for independent events, to calculate the probability that variable \(A\) and \(B\) as the \(P(A\, \text{and}\, B) = P(A) \cdot P(B)\). Remember, in a hypothesis test, you assume that \(H_{0}\) is true, and the two variables are assumed to be independent. \[\begin{aligned} P(A\, \text{and}\, B) &= P(A) \cdot P(B) \text{ if } A \text{ and } B \text{ are independent} \\[4pt] &=\dfrac{\text { Number of ways A can happen }}{\text { Total number of individuals }} \cdot \dfrac{\text { Number of ways B can happen }}{\text { Total number of individuals }} \\[4pt] &=\dfrac{\text { Row Total }}{n} \cdot \dfrac{\text { Column Total }}{n} \end{aligned}\]
To compute the expected frequency for each cell in the contingency table, perform the following steps. Find out how many individuals you expect to be in a certain cell. To find the expected frequencies, you just need to multiply the probability of that cell times the total number of individuals. Do not round the expected frequencies. =\[\begin{aligned} \text{Expected frequency (cell A and B)} &= E(A \text{ and } B) \\ &= n \left(\dfrac{\text { Row Total }}{n} \cdot \dfrac{\text { Column Total }}{n}\right) = \dfrac{\text { Row Total } \cdot \text { Column Total }}{n} \end{aligned}\] If the variables are independent, the expected frequencies and the observed frequencies should be the same.
The test statistic here will involve looking at the difference between the expected frequency and the observed frequency for each cell. Then you want to find the “total difference” of all of these differences. The larger the total, the smaller the chances that you could find that test statistic given that the assumption of independence is true. That means that the assumption of independence is not true.
How is the test statistic calculated? First, compute the differences between the observed and expected frequencies. Since some of the differences may be positive and others negative, they are squared to ensure all values are positive and to emphasize larger deviations. These squared differences might be large simply because the frequencies themselves are large, so each one is divided by its corresponding expected frequency to scale the values appropriately. Afterward, all of these scaled values are added together. This process measures the overall variance between observed and expected frequencies. The result is compared to a chi-square distribution to determine the critical value or p-value. For this reason, the procedure is known as a chi-square test.
The \(\chi^{2}\)-test is a statistical test for testing the independence between two variables. It can be used when the data are obtained from a random sample, and when the expected value \((E)\) from each cell is 5 or more.
- The test statistic is: \(\chi^{2} = \sum \frac{(O-E)^{2}}{E}\)
- Use the table with degrees of freedom and \(\alpha\).
- The degrees of freedom are \(\text{df}\) = (the number of rows – 1) (the number of columns – 1), that is, \(\text{df} = (R-1)(C-1)\)
- The observed frequency (sample results) is denoted as \(O\)
- The expected frequency is denoted as \(E\)
Steps to Perform the
- Identify the claim and state the hypotheses.
- Find the critical value using the table with df = (the number of rows – 1) (the number of columns – 1) and \(\alpha\).
- Compute the test statistic using \(\chi^{2} = \sum \frac{(O-E)^{2}}{E}\).
- Decide whether to reject or not reject the null hypothesis (H0).
- Summarize the results.
Examples of
Is there an association between autism spectrum disorder (ASD) and breastfeeding? To explore this, a researcher surveyed mothers of children with and without ASD about the duration of time they breastfed their children. Does the data provide sufficient evidence to determine whether breastfeeding and ASD status are independent? Conduct the test using an \(\alpha\) = 0.01.
| ASD | None | Less than 2 months | 2 to 6 months | Over 6 months | Total |
|---|---|---|---|---|---|
| Yes | 241 | 198 | 164 | 215 | 818 |
| No | 20 | 25 | 27 | 44 | 116 |
| Total | 261 | 223 | 191 | 259 | 934 |
Table \(\PageIndex{1}\): Length of Breast Feeding and ASD Status
(Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006.)
Solution
- Identify the claim and state the hypotheses.
\(H_{0}:\) Autism spectrum disorder and length of breastfeeding are independent (Claim)
\(H_{1}:\) Autism spectrum disorder and length of breastfeeding are dependent
- Find the critical value using the table.
In this problem, there are 2 rows and 4 columns. Thus, df = (2 – 1)(4 – 1) = 1 \(\cdot\) 3 = 3. Using df = 3 and \(\alpha\) = 0.01, the critical value is 11.345.
- Compute the test statistic.
Compute the expected values for each cell using the observed value in the table above using \(\text { Expected Value } = \dfrac{\text { Row Total } \cdot \text { Column Total }}{\text { Grand Total }}\).
| ASD | None | Less Than 2 Months | 2 to 6 Months | Over 6 Months | Total |
|---|---|---|---|---|---|
| Yes | \(\dfrac{818 \cdot 261}{934}\)= 228.585 | \(\dfrac{818 \cdot 223}{934}\) = 195.304 | = 167.278 | \(\dfrac{818 \cdot 259}{934}\) = 226.833 | 818 |
| No | \(\dfrac{116 \cdot 261}{934}\) = 32.415 | \(\dfrac{116 \cdot 223}{934}\) = 27.696 | \(\dfrac{116 \cdot 191}{934}\) = 23.722 | \(\dfrac{116 \cdot 259}{934}\) = 32.167 | 116 |
| Total | 261 | 223 | 191 | 259 | 934 |
Table \(\PageIndex{2}\): Computation of Expected Values Per Cell
It will be helpful to make a table for the expected counts and another one for each of the \(\frac{(O-E)^{2}}{E}\) values to aid in computing the test statistic.
| ASD | None | Less Than 2 Months | 2 to 6 Months | Over 6 Months | Total |
|---|---|---|---|---|---|
| Yes | \(\dfrac{(241-228.585)^2}{228.585}\) = 0.6743 | \(\dfrac{(198-195.304)^2}{195.304}\) = 0.0372 | \(\dfrac{(164-167.278)^2}{167.278}\) = 0.0642 | \(\dfrac{(215-226.833)^2}{226.833}\) = 0.6173 | 818 |
| No | \(\dfrac{(20-32.415)^2}{32.415}\) = 4.7552 | \(\dfrac{(25-27.696)^2}{27.696}\) = 0.2624 | \(\dfrac{(27-23.722)^2}{23.722}\) = 0.4531 | \(\dfrac{(44-32.167)^2}{32.167}\) = 4.3529 | 116 |
| Total | 261 | 223 | 191 | 259 | 934 |
Table \(\PageIndex{3}\): Computation of \(\frac{(O-E)^{2}}{E}\) Per Cell
The test statistic is the sum of all eight \(\frac{(O-E)^{2}}{E}\) values: \(\chi^{2}=\sum \frac{(O-E)^{2}}{E} = 11.217\).
- Decide whether to reject or not reject the null hypothesis (H0).
Since the test point falls into the non-critical region, the result is not to reject the null hypothesis (H0).
- Summarize the results.
There is not enough evidence to show a relationship between autism spectrum disorder and breastfeeding.
A community college researcher in southern California wants to know: Is there a dependent relationship between a student’s gender and the type of school they plan to transfer to? To find out, the counselor surveys 800 students, recording their gender (male or female) and the type of four-year school they intend to transfer to: UC (University of California), Cal State, Out-of-State, or Private. Test the researcher's claim using the p-value method with an\(\alpha\) = 0.05. The observed values are provided in the contingency table below.
| UC | Cal State | Out - of - State | Private | Total | |
|---|---|---|---|---|---|
| Male | 116 | 168 | 63 | 47 | 394 |
| Female | 104 | 132 | 77 | 93 | 406 |
| Total | 220 | 300 | 140 | 140 | 800 |
Table \(\PageIndex{4}\): Student's Gender and Transfer School Type
Solution
- Identify the claim and state the hypotheses.
H₀: Gender and preferred transfer school are independent
H₁: Gender and preferred transfer school are not independent (Claim)
- Compute the test statistic and p-value using a TI-84+ calculator.
- Enter the Observed Data into a Matrix
- Press [2nd] then [X–1] to open the MATRIX menu.
- Scroll right to [EDIT], then select [A] by pressing [1].
- Enter the matrix dimensions as 2 × 4 (2 rows for gender, 4 columns for school type).
- Enter the observed values row by row: Row 1 (Male): 116, 168, 63, 47, and Row 2 (Female): 104, 132, 77, 93.]
- Enter the Observed Data into a Matrix
- Perform the \(\chi^2\) test.
- Press [STAT] and scroll right to TESTS.
- Scroll down and select C:\(\chi^2\) Test.
- Set the following: Observed Matrix: [A] and Expected Matrix: [B] (the calculator will compute this).
- Scroll to Calculate and press [ENTER].
- Read the results.
- The test point is \(\chi^2\) = 21.314 (rounded to three place values)
- The p-value = 9.061E-5, which is rewritten by moving the decimal places 5 places to the left to get 0.000091.
- Decide whether to reject or not reject the null hypothesis (H0).
Since the p-value = 0.000091 < \(\alpha\) = 0.05, the result is to reject the null hypothesis.
- Summarize the results.
There is enough evidence to support the claim that gender and preferred transfer school are related.
A researcher is curious whether the size of a company influences whether it offers dental insurance to its employees. To explore this, data were collected from a sample of small, medium, and large companies, recording how many in each category provide dental coverage. Use an \(\alpha\) = 0.10.
Based on the results, is there evidence of an independent relationship between company size and dental insurance coverage?
Use the data below to perform a chi-square test for independence and determine whether the type of company is associated with offering dental benefits.
| Dental Insurance: | Small Company | Medium Company | Large Company | Total |
|---|---|---|---|---|
| Yes | 21 | 25 | 19 | 65 |
| No | 46 | 39 | 10 | 95 |
| Total | 67 | 64 | 29 | 160 |
Table \(\PageIndex{5}\): Size of Company and Dental Insurance
Solution
- Identify the claim and state the hypotheses.
\(H_{0}:\) Dental insurance coverage and company size are independent (claim)
\(H_{1}:\) Dental insurance coverage and company size are dependent
- Find the critical value using the table.
In this problem, there are 2 rows and 3 columns. Thus, df = (2 – 1)(3 – 1) = 1 \(\cdot\) 2 = 2. Using df = 2 and \(\alpha\) = 0.10, the critical value is 4.605.
- Compute the test statistic.
Compute the expected values for each cell using the observed value in the table above using \(\text { Expected Value } = \dfrac{\text { Row Total } \cdot \text { Column Total }}{\text { Grand Total }}\).
| Dental Insurance: | Small Company | Medium Company | large Company | Total |
|---|---|---|---|---|
| Yes | \(\dfrac{65 \cdot 67}{160}\)= 27.219 | \(\dfrac{65 \cdot 64}{160}\) = 26 | \(\dfrac{65 \cdot 29}{160}\) = 11.781 | 65 |
| No | \(\dfrac{95 \cdot 67}{160}\) = 39.781 | \(\dfrac{95 \cdot 64}{160}\) = 38 | \(\dfrac{95 \cdot 29}{160}\) = 17.219 | 95 |
| Total | 67 | 64 | 29 | 160 |
Table \(\PageIndex{6}\): Computation of Expected Values Per Cell
It will be helpful to make a table for the expected counts and another one for each of the \(\frac{(O-E)^{2}}{E}\) values to aid in computing the test statistic.
| Dental Insurance: | Small Company | Medium Company | Large Company | Total |
|---|---|---|---|---|
| Yes | \(\dfrac{(21-27.219)^2}{27.219}\) = 1.421 | \(\dfrac{(25-26)^2}{26}\) = 0.038 | \(\dfrac{(19-11.781)^2}{11.781}\) = 4.424 | 65 |
| No | \(\dfrac{(46-39.781)^2}{39.781}\) = 0.972 | \(\dfrac{(39-38)^2}{38}\) = 0.026 | \(\dfrac{(10-17.219)^2}{17.219}\) = 3.027 | 95 |
| Total | 67 | 64 | 29 | 160 |
Table \(\PageIndex{7}\): Computation of \(\frac{(O-E)^{2}}{E}\) Values Per Cell
The test statistic is the sum of all six \(\frac{(O-E)^{2}}{E}\) values: \(\chi^{2}=\sum \frac{(O-E)^{2}}{E} = 9.908\).
Remember that the TI-84+ calculator can be used to find both the test statistic and the p-value. The results are shown below. These values are more accurate than those from manual calculations because the calculator avoids rounding errors.
- Decide whether to reject or not reject the null hypothesis (H0).
- Traditional method: since the test point = 9.908 > Critical Value = 4.605, reject the null hypothesis (H0).
- P-value Method: since the p-value = 0.0071 < \(\alpha\) = 0.10, reject the null hypothesis (H0).
- Summarize the results.
There is enough evidence to support the claim that there is a relationship between dental insurance coverage and company size.
Authors
"11.2: Test for Independence" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0
Attributions
"11.1: Chi-Square Test for Independence" by Kathryn Kozak is licensed under CC BY-SA 4.0
"10.3: Test for Independence" by Rachel Webb is licensed under CC BY-SA 4.0
Exercises
- The number of people who survived the Titanic based on class and gender ("Encyclopedia Titanica," 2013). Is there enough evidence to show that the class and the gender of a person who survived the Titanic are independent? Test at the 5% level. Use the traditional method.
| Class | Female | Male | Total |
|---|---|---|---|
| 1st | 134 | 59 | 193 |
| 2nd | 94 | 25 | 119 |
| 3rd | 80 | 58 | 138 |
| Total | 308 | 142 | 450 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- The number of people who survived the Titanic based on class and gender ("Encyclopedia Titanica," 2013). Is there enough evidence to show that the class and the gender of a person who survived the Titanic are independent? Test at the 5% level. Use the p-value method.
| Class | Female | Male | Total |
|---|---|---|---|
| 1st | 134 | 59 | 193 |
| 2nd | 94 | 25 | 119 |
| 3rd | 80 | 58 | 138 |
| Total | 308 | 142 | 450 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- Is there a relationship between autism and what an infant is fed? To determine if there is, a researcher asked mothers of autistic and non-autistic children to say what they fed their infants. The data is in Example 11.2.7 (Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough evidence to show dependence on what an infant is fed and autism? Test at the 1% level. Use the traditional method.
| Autism | Breast Feeding | Formula with DHA/ARA | Formula without DHA/ARA | Row Total |
|---|---|---|---|---|
| Yes | 12 | 39 | 65 | 116 |
| No | 6 | 22 | 10 | 38 |
| Column Total | 18 | 61 | 75 | 154 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- Is there a relationship between autism and what an infant is fed? To determine if there is, a researcher asked mothers of autistic and non-autistic children to say what they fed their infants. The data is in Example 11.2.7 (Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough evidence to show dependence on what an infant is fed and autism? Test at the 1% level. Use the p-value method.
| Autism | Breast Feeding | Formula with DHA/ARA | Formula without DHA/ARA | Row Total |
|---|---|---|---|---|
| Yes | 12 | 39 | 65 | 116 |
| No | 6 | 22 | 10 | 38 |
| Column Total | 18 | 61 | 75 | 154 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- Students at multiple grade schools were asked what their personal goals (get good grades, be popular, be good at sports) were and how important having good looks was to them (1 very important and 4 least important). The data is in Example 11.2.11 ("Popular kids datafile," 2013). Does the data provide enough evidence to show that goal attainment and the importance of looks are independent? Test at the 5% level. Use the traditional method.
| Goals | 1 | 2 | 3 | 4 | Row Total |
|---|---|---|---|---|---|
| Grades | 80 | 66 | 66 | 35 | 247 |
| Popular | 81 | 30 | 18 | 12 | 141 |
| Sports | 24 | 30 | 17 | 19 | 90 |
| Column Total | 185 | 126 | 101 | 66 | 478 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- Students at multiple grade schools were asked what their personal goals (get good grades, be popular, be good at sports) were and how important having good looks was to them (1 very important and 4 least important). The data is in Example 11.2.11 ("Popular kids datafile," 2013). Does the data provide enough evidence to show that goal attainment and the importance of looks are independent? Test at the 5% level. Use the p-value method.
| Goals | 1 | 2 | 3 | 4 | Row Total |
|---|---|---|---|---|---|
| Grades | 80 | 66 | 66 | 35 | 247 |
| Popular | 81 | 30 | 18 | 12 | 141 |
| Sports | 24 | 30 | 17 | 19 | 90 |
| Column Total | 185 | 126 | 101 | 66 | 478 |
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.
- Answers
-
If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.