Questions and Answers #2 Chi-squared test

Questions and Answers Sheet 2 Chi-squared test Question #1 For a goodness-of-fit test with 𝛼 = 0.10 and 5 categories, what is the critical χ2 value? Answer: Obtaining the critical value for chi-square test: Here, the level of significance for the test is, α=0.10 and the degrees of freedom is 4 (number of categories –1). Critical value for left tailed test: The critical value for a left tailed chi-square test at 𝛼 = 0.10 is obtained using the EXCEL formula, “=CHISQ.INV (0.10, 4)”. From Excel, the critical value is 1.0636. Thus, the critical value for the left tailed chi-square test is 1.0636. Critical value for right tailed test: The critical value for a right tailed chi-square test at 𝛼 = 0.10 is obtained using the EXCEL formula, “=CHISQ.INV.RT (0.10, 4)”. From Excel, the critical value is 7.7794. Thus, the critical value for the right tailed chi-square test is 7.7794. Critical value for two tailed test: The critical value for a two tailed chi-square test at 𝛼 = 0.10 is obtained using the EXCEL formula, “=CHISQ.INV (0.95, 4)”. From Excel, the critical value is 9.4877. Thus, the critical value for the two tailed chi-square test is 9.4877. Answer: The critical value for the left tailed chi-square test is 1.0636. The critical value for the right tailed chi-square test is 7.7794. The critical value for the two tailed chi-square test is 9.4877. Question #2 I need answer full parts to the question. Show all work and be neat. Show all conditions, and p-value, conclusion, anything under the 4 step statistical process. It can be difficult to motivate preschoolers to practice writing letters. Perhaps a reward system will do the trick, but what type of rewards work? Here are the data from a randomized experiment that compared three treatments among children in preschool. A “success" means that the child completed a worksheet in which they traced the entire alphabet in one sitting. Group Reward Subjects Successes 1 Choice of a small toy 20 13 ( 2 ) Choice of candy 24 18 3 Promise of immediate recess 16 10 (a) Which chi-square test is appropriate in this setting? Explain your reasoning (b) State the appropriate hypotheses for performing a chi-square test in this setting. (c) The chi-square test statistic is 𝜒 2 = 0.847 . Calculate the P-value. (d) Interpret the P-value (e) What conclusion would you draw? Answer: Hypothesis testing is conducted for the parameters of the population like population mean, population standard deviation and so on. It is to check whether the sample values conform to the prestated parameter values of the original distribution or not. There are various tests that are used while conducting the hypothesis testing. These are: 1.z test 2.t test 3.F test 4.Chi-square test. The chi square test could be classified as: 1.Chi-square test for independence. 2.Chi-square test for goodness. 3.Chi-square test for homogeneity. The Chi-Square Test of Independence is mainly used to check whether there is an association between categorical or qualitative variables, that is, whether the variables are independent or related (dependent) or not. The formula to compute the test statistic for chi-square test of independence is: 𝜒2 = ∑ ( 𝑂−𝐸 2 𝐸 ) Here, 𝜒 2 = Chi square test statistic 𝑂 = Observed frequency 𝐸 = Expected frequency ∑ = Summation (a) In this case, two different categorical variables are being measured. Hence, the chi-square test of independence is appropriate here. (b) The hypotheses are: 𝐻0 :There exists no relationship between two groups 𝐻𝑎 :There exists a relationship between two groups (c)The degree of freedom is: df=Number of categories -1 =3-1=2 The P-value is 0.655. (d) Here, P-value is higher than assumed significance level. We cannot reject the null hypothesis. (e) There is sufficient proof to say that the two variables are not dependent. Question #3 If the chi-square test is 𝜒 2 (4, 𝑁 = 71) = 10, if the alpha is .05, then we ... Answer: According to the provided data, the chi-square test statistic is 10. The sample size, N is 71 and the degrees of freedom, df = 4. The level of significance, α is 0.05. The P-value can be obtained as: Using chi-square distribution table for upper-tail areas, the P-value is obtained as 0.0404. It is obtained by checking the df equal to 4 in row and the value of chi-square test statistic, 10 in that particular column. So, the corresponding right-tail area is the required P-value. Since, P-value ≤ 𝛼 , that is, 0.0404 ≤ 0.05 , then we reject the null hypothesis. Question #4 In the Salsberry (2003) study shared in this weeks module, the researcher presents the following results of the statistical analysis: "With respect to family configuration, 20% of the children enrolled in Medicaid, 29% of the uninsured children, and 44% of the privately insured children were in families where the adults were in a “partnered relationship”, [𝜒 2 (2, 𝑁 = 392) = 21.95; 𝑃 = .0001] " What type of analysis does this statement reflect? Answer: Chi-square test for “Goodness of fit”: The Chi-square test for “Goodness of fit” is used to test the shape or proportion of the population distribution. Moreover, the percentage or proportion for each category is provided in the Chi-square test for “Goodness of fit”. The general formula for the Chi-square test for “Goodness of fit” is, 𝜒2 = ∑ (𝑂−𝐸)2 𝐸 with C-1 degrees od freedom Where O is observed frequency, E is expected frequency and C is number of Category From the statement, it is observed that the percentage of family configuration is given. Moreover, the value of 𝜒 2 and p-value is provided. Hence, the analysis that the statement reflect is Chi-square test for “Goodness of fit” Question #5 IQ scores among the general population have a mean of 100 and a standard deviation of 15. A researcher claims that the standard deviation, o, of IQ scores for males is less than 15. A random sample of 19 IQ scores for males had a mean of 101 and a standard deviation of 10. Assuming that IQ scores for males are approximately normally distributed, is there significant evidence (at the 0.05 level of significance) to conclude that the researcher's claim is correct? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis 𝐻0 , and the alternative hypothesis 𝐻1 , 𝐻0 : 𝐻1 : (b) Determine the type of test statistic to use. (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the critical value. (Round to three or more decimal places.) (e) Can we support the claim that the standard deviation of IQ scores for males is less than 15? a. Yes b. No Answer: The chi-square test for variance is a non-parametric statistical procedure with a chi-square-distributed test statistic that is used for determining whether the variance of a variable obtained from a particular sample has the same size as the known population variance of the same variables. Given data is appropriate for testing of hypothesis to test chi-square for single variance. Because it is a small sample test. a) Null Hypothesis: 𝐻0 : 𝜎 = 15 Alternative Hypothesis 𝐻1 : 𝜎 > 15 Level of significance=5% b) The type of test statistic is chi-square test for single variance c) Test statistic 𝜒 2 = (𝑛 − 1)𝑠 2 /𝜎 2 = (19 − 1)(102 )/(152 ) = (18 × 100)/225 =8 d) Critical value Read the chi square table value at 5% level of significance for right tailed test at 18 degrees of freedom is 2 𝜒𝑡𝑎𝑏 = 9.390 e) Yes the claim the standard deviation of IQ scores for males is less than 15. Question #6 Let 𝑝1 = population proportion for population 1, 𝑝2 = population proportion for population 2, ... and 𝑝𝑘 = population proportion for population k. Consider the following null hypothesis: 𝐻0 : 𝑝1 = 𝑝2 = ⋯ = 𝑝𝑘 . Which of the following statements is correct? a. The alternative hypothesis to the null hypothesis stated above must be: 𝐻𝛼 : Not all population proportions are equal. b. If the sample data and the chi-square test computations indicate 𝐻0 cannot be rejected, we cannot detect a difference among the k population proportions. c. If the sample data and the chi-square test computations indicate 𝐻0 can be rejected, we have the statistical evidence to conclude that one or more population proportions differ from the other population proportions. d. All of the above. Answer: Option d) is correct. All of the above. a) The alternative hypothesis to the null hypothesis stated above must be: 𝐻𝛼 : Not all population proportions are equal. b) If the sample data and the chi-square test computations indicate 𝐻0 cannot be rejected, we cannot detect a difference among the k population proportions. c) If the sample data and the chi-square test computations indicate 𝐻0 can be rejected, we have the statistical evidence to conclude that one or more population proportions differ from the other population proportions. Question #7 Observed and Expected counts are given for a chi-square test for association, with the Expected counts in parentheses. Calculate the chi-square statistic for this test. A B C 1 33(40.4) 34(31.6) 33(28) (2 22(40.4) 51(31.6) 27(28)) 3 147(121.2) 73(94.8) 80(84) Round your answer to three decimal places. 𝜒2 = Answer: Chi-square test: The chi-square (𝜒 2 )statistic is a test that measures how a model compares to actual observed data. The data used in calculating a chi-square statistic must be random. the chi-square (𝜒 2 ) test is used for the hypothesis testing. There are two types of hypothesis tests. 1) A chi-square goodness of fit test: this test determines if sample data matches a population. 2) A chi-square test for independence: this test is used for compares two variables in a contingency table to see if they are related or not. The test statistics can be calculated as, 𝜒2 = ∑ (𝑂𝑖 − 𝐸𝑖 )2 = 33.452 𝐸𝑖 Therefore, the test statistic is 33.452. Excel output: Observed value (O) 33 34 33 22 51 ( 27 147 73 80 Total Expected value (E) 40.4 31.6 28 40.4 31.6 28 121.2 94.8 84 (O-E) / E 1.355445545 0.182278481 0.892857143 8.38019802 11.91012658 ) 0.035714286 5.492079208 5.013080169 0.19047619 33.45225562 Question #8 Physical therapy (PT) may be just as successful as traditional surgery when treating a specific type of knee ligament tear in elderly patients. A group of researchers at a large teaching hospital conduct a trial in which patients are randomly assigned to one of three treatments. A "success" means that the subject had regained full range of motion in their knee 6 months post- treatment. Surgery ( PT Surgery+PT Success 56 54 85 Failure 22 35 ) 18 When analyzing their data, which test would be appropriate for these researchers to use? Chi-square test of/for Answer: The chi-square test of homogeneitytests to see whether different columns (or rows) of data in a table come from the same population or not (i.e., whether the differences are consistent with being explained by sampling error alone). Answer : The test would be appropriate for these researchers to use is : Chi-square test of homogeneity. Question #9 Random-digit-dialing telephone surveys used to exclude cell phone numbers. If the opinions of people who have only cell phones differed from those of people who have landline service, the poll results may not represent the entire adult population. The Pew Research Center interviewed separate random samples of cell-only and landline telephone users who were less than 30 years old and asked them to describe their political party affiliation. or goodness of fit, homogeneity or independence? Answer: Given, Random-digit-dialing telephone surveys used to exclude cell-phone numbers. If the opinions of people who have only cell phones differ from those of people who have landline service, the poll results may not represent the entire adult population. The Pew Research Center interviewed separate random samples of cell-only and landline telephone users who were less than 30 years old and asked them to describe their political party affiliation. When we are interested in the distribution of one variable, then we will use a chi-square goodness-of-fit test. When we are interested in the distribution of two variables in which there are multiple independent samples, then in this scenario we have to use a chi-square test for homogeneity. When we are interested in the distribution of two variables and if there is a single sample, then we would like to use a chi-square test for independence. In this scenario, there are two variables. Type of telephone and political party. There are two random samples of cell-only and landline telephone users. Therefore, we should use Chi-square test for homogeneity. Question #10 Suppose that you are interested in examining the relationship between executive compensation and three independent variables that you believe have an effect: Proficiency in a second language, education level, and work experience abroad. You want to be sure, however, that conditional heteroscedasticity is not present in your data. Therefore, you regress the squared residuals of the regression model onto the three independent variables, and you find that the 𝑅2 for the regression 0.181. Your sample size of executives is 36. a. What is the name of the test that is used to determine if conditional heteroscedasticity is present? b. Using the information provided, calculate the F-statistic that you would use to determine if there is conditional heteroskedasticity present. c. Using the information provided, calculate the 𝜒 2 test statistic (LM method) that you would use to determine if there is conditional heteroskedasticity present. d. What critical value would you use if you are testing at a 5% significance level using the F-test approach? e. What critical value would you use if you are testing at a 5% significance level using the 𝜒 2 test approach? Answer: Introduction F-static 𝑅2 (𝑘 − 1) 𝐹= (1 − 𝑅2 ) (𝑛 − 𝑘) 𝐹= 𝜒2 𝑎 where a=degree of freedom Explanation a) The name of the test that is used to determine if conditional heteroscedasticity is present is chisquared test . b) n=36 k=3 𝑅2 = 0.181 𝐹= 𝐹= 0.181 (3−1) (1−0.181) (36−3) 0.905 0.273 = 0.331 c) chi-squared test F-static=0.331 Degree of freedom of chi-squared test =(n-k)=(36-3)=33 Chi -squared =33(0.331)=10.939 d) F-statistic p-value at 5% level of significance p-value =0.7205 (calculated by using p-value calculator for f-test) since p-value is less critical region therefore, null hypothesis is reject at 5% level of significance .So , the test is significant. Conclusion a)The name of the test that is used to determine if conditional heteroscedasticity is present is chi-squared test . b) f-static =0.331 c) chi squared test =10.939 d) F-test Test is significant at 5% level of significance. Question #11 Suppose we observe the following random sample of size n = 117, sampled from a discrete distribution. Value ( Count 0 1 2 3 4 5 6 32 36 21 18 6 2 2) We would like to test whether this data comes from a Poisson( 𝜆 ) distribution. calculate the observed value of the appropriate Goodness of Fit test statistic, used to test the null hypothesis that the observed data is Poisson distributed. Give answer to three decimal places. Compute the p-value of the observed test statistic you just computed. Give answer to three decimal places. Answer: Chi-Square Test: In statistical hypothesis test, a chi-square test, or specifically Pearson's chi-square test is performed when the test statistic is distributed as chi-square under the null hypothesis. This test is used to determine whether there is a significant difference between the observed and the expected frequencies in one or more categories of a contingency table. In the standard applications of this test, the observations are classified into mutually exclusive classes. If the null hypothesis is true i.e. there are no significant differences between the classes in the population, the test statistic which is computed from the observations follows a chi-square frequency distribution. The purpose of the test is to determine how likely the observed frequencies would be assuming the null hypothesis is true. If 𝑂𝑖 and 𝐸𝑖 be the observed and expected frequency for the 𝑖 𝑡ℎ category, then the chi-square test statistic is given by, 𝜒𝑐2 = ∑ 𝑖 (𝑂𝑖 − 𝐸𝑖 )2 𝐸𝑖 where c is the degrees of freedom of the chi-square distribution. Here, the null hypothesis is that the data comes from a Poisson distribution. So, from the given data, we find the mean of the values. Value (x) 0 1 2 3 ( 4 5 6 Total Count (f) 32 36 21 18 6 2 2 117 xf 0 36 42 54 ) 24 10 12 178 Therefore, 1 178 𝑁 117 𝜒 = ∑𝑛𝑖=1 𝜒𝑖 𝑓𝑖 = = 1.52 (rounded to 2 decimal places) Now, the parameter of the population is 𝜆 i.e. Average = 𝜆 = 1.52 Then the probability mass function of the random variable of the value is given by, (𝑃(𝑋 = 𝑥) = ( 𝑒 −1.52 (1.52) 𝜒 ) , (0): ), 𝑓𝑜𝑟𝜒 = 0,1,2, … ∞ 𝜒! Using this, we calculate the expected frequencies multiplying each probability by the total frequency i.e. 117. Value (x) 0 1 2 3 ( 4 5 6 Total We have, Observed frequency (Oi ) 32 36 21 18 6 2 2 117 Expected frequency ( Ei) (Oi − Ei ) 2 / Ei 25.6 1.6060 38.9 0.2156 29.6 2.4792 15.0 0.6100 ) 5.7 0.0167 1.7 0.0421 0.4 5.5642 117 10.5337 𝜒𝑐2 = ∑𝑖 (𝑂𝑖 −𝐸𝑖 )2 𝐸𝑖 =10.5337 (rounded to 4 decimal places) The value of the test statistic is 10.534 (rounded to 3 decimal places). From the chi-square value, we calculate p-value for this test using MS-Excel, p-value=CHISQ.DIST(chi-square value,n-1,TRUE) =CHISQ.DIST(10.534,6,TRUE) =0.896(rounded to 3 decimal places) Answer: The p-value of the observed test statistic is 0.896. Question #12 The _(1)_ are the data for a Chi-Square test, and you get them by simply counting. The _(2)_ are something you get from either past research, or an assumption that all categories are equal, and they go in the null hypothesis. The _(3)_ are something you get by multiplying sample size, n, by expected proportions. Group of answer choices: (1) observed frequencies, 𝑓𝑜 (2) expected proportions (3) expected frequencies, 𝑓𝑒 (1) expected frequencies, 𝑓𝑒 (2) expected proportions (3) observed frequencies, 𝑓𝑜 (1) expected proportions (2) observed frequencies, 𝑓𝑜 (3) expected frequencies, 𝑓𝑒 (1) expected frequencies, 𝑓𝑒 (2) observed frequencies, 𝑓𝑜 (3) expected proportions Answer: The study is about the chi-square test The chi-square is a non-parametric test which is used for comparing the categorical variables. The chisquare goodness of fit test helps in determining how the observed distribution fits the expected distribution. It is used for comparing the observed data with the expected data. The observed frequencies, 𝑓𝑜 are the data for a Chi-Square test, and you get them by simply counting. The expected proportions are something you get from either past research, or an assumption that all categories are equal, and they go in the null hypothesis. The expected frequencies, 𝑓𝑒 are something you get by multiplying sample size, n, by expected proportions. Correct Answers: (1) observed frequencies, 𝑓𝑜 (2) expected proportions (3) expected frequencies, 𝑓𝑒 . Question #13 The Bitter Bottling Company has developed "Featherweight," the cola with fewer calories and less taste. To evaluate this new product, the marketing manager gives a taste test to a random sample of 300 people. Each person in the sample tastes Featherweight and 4 other diet cola brands. To avoid bias, the actual brand labels are replaced by the letters A, B, C, D, and E. The results of the sample are shown in the following table: Brand A B ( C D E Number of Preferring Brand 50 65 ) 45 70 70 Do the customers prefer a particular brand more than the others? Use an alpha of 0.05. 1.What type of hypothesis testing should be used? a. Chi-square Goodness of Fit Test b. Chi-square Test on Independence 2. What are the null and alternative hypothesis? a. 𝐻0 : The market does not have preferences; 𝐻1 : The market does have preference b. 𝐻0 : The market does have preferences; 𝐻1 : The market does not have preference 3. What are the significance level and the type of test? a. 𝛼 = 0.05 , Right-tailed b. 𝛼 = 0.05 , left-tailed c. 𝛼 = 0.05 , two-tailed Answer: The table shows the number of preferring brand with actual brands labels letters A, B, C, D, and E. 1. Chi-square Goodness of fit test: The chi-square is a non-parametric test which is used for comparing the categorical variables. The chi-square goodness of fit test helps in determining how the observed distribution fits the expected distribution. It is used for comparing the observed data with the expected data. The variables Brands and observation group(Number preferring Brand) are categorical. The chi-square goodness of fit test can be used for testing the customers prefer a particular brand more than the other using the expected frequencies. Thus, the type of hypothesis testing should be used is Chi-square Goodness of Fit Test. Correct Answer: Option 1. 2. The hypothesis is, Null hypothesis: 𝐻0 : The market does not have preferences. Alternative hypothesis: 𝐻1 : The market does have preferences. Correct Answer: Option 1. Question #14 The manager of a computer network has collected data on the number of times that service has been interrupted on each day over the past 500 days. The results are as follows as shown in the table. Does this distribution of service interruptions follow the expected percentages which are 0=30%, 1=30%, 2= 20%, 3=12%, 4=3%, 5=3%, and 6=2%? Use the 0.01 level of significance. Interruptions per Day 0 1 2 ( 3 4 5 6 Number of Days 160 175 86 ) 41 18 12 8 1. What type of hypothesis testing should be used? a. Chi-square Goodness of Fit Test b. Chi-square Test on Independence 2. What are the null and alternative hypothesis? a. 𝐻0 : The observed frequency does not follow the expected distribution; 𝐻1 : the observed frequency follows the expected distribution b. 𝐻0 : The observed frequency follows the expected distribution; 𝐻1 : The observed frequency does not follow the expected distribution 3. What are the significance level and the type of test? a. 𝛼 = 0.01 , Right-tailed b. 𝛼 = 0.01 , left-tailed c. 𝛼 = 0.01 , two-tailed Answer: The data shows the number of days that the services has been interrupted. 1. Chi-square Goodness of fit test: The chi-square is a non-parametric test which is used for comparing the categorical variables. The chi-square goodness of fit test helps in determining how the observed distribution fits the expected distribution. It is used for comparing the observed data with the expected data. The variables number of interrupts per day and number of days are categorical. The chi-square goodness of fit test can be used for testing the distribution of service interruptions by using the expected percentages. Thus, the type of hypothesis testing should be used is Chi-square Goodness of Fit Test. Correct Answer: a. 2. The hypothesis is, Null hypothesis: 𝐻0 : The observed frequency follows the expected distribution. Alternative hypothesis: 𝐻1 : The observed frequency does not follow the expected distribution. Correct Answer: b. Question #15 Use Table VII to find the required 𝜒 2 -values. Illustrate your work graphically. When you use chi-square procedures to make inferences about a population standard deviation, why should the variable under consideration be normally distributed or nearly so? Answer: A Chi-square distribution is the sum of squares of the standard normal deviates. The Chi-square test can be used for checking the association between categorical variables. The assumptions for the chi-square test are: 1. The sampling method should be simple random sampling (normality). 2. The variables should be categorical in nature. 3. The expected value of the number of sample observations inn each level of the variable is at least 5. The one standard deviation chi-square test is not robust to moderate violations of the normality assumption, unlike the z-tests and t-tests for one and two population means. It is used only when there is considerable evidence that the variable under consideration is normally distributed or nearly so. Question #16 What is the degrees of freedom in the follow statistical statement 𝜒 2 (2, 𝑁 = 67) = 6.72, 𝑝 = .021 ? Answer: The Chi-square test is a test for categorical variables which determines whether there is a difference between population proportions between two or more groups. Consider this case. The value of Chi-square test statistic is given as: 𝜒 2 (2, 𝑁 = 67) = 6.72 Where N is the sample size, 2 is degrees of freedom, and 6.72 is the chi-square test statistic. The degrees of freedom in the follow statistical statement are 2. Question #17 You are conducting a Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Report all answers correct to three decimal places. Category A B ( C D E Observed Frequency 15 19 23 18 5 Expected Frequency What is the chi-square test-statistic for this data? 𝜒2 = At the alpha = 0.05 level, what is the conclusion for this test? Fail to reject the null hypothesis Reject the null hypothesis Report all answers accurate to three decimal places. Answer: The hypotheses are given below: Null hypothesis: 𝐻0 : All five categories are equally likely to be selected. Alternative hypothesis: 𝐻1 : All five categories are not equally likely to be selected. The chi-square test is appropriate in this situation:” The test statistic is given below: 𝜒2 = ∑ (𝑂𝑖 −𝐸𝑖 )2 𝐸𝑖 2 ∼ 𝜒(𝑘=1) Where 𝑂𝑖 Observed frequency 𝐸𝑖 Expected frequency k=Number of observations (obs-exp) 2 /exp ) The expected frequency is obtained as given below: 𝐸= ∑Observed 𝑛 = 80 5 = 16 Test statistic Category A B ( C D E Observed Frequency 15 19 23 18 5 (obs-exp) 2 /exp 0.063 0.563 ) 3.063 0.25 7.563 Expected Frequency 16 16 16 16 16 The chi-square test-statistic for this data: 𝜒 2 = 0.063 + 0.563 + 3.063 + 0.25 + 7.563 = 11.5. P-value: The P-value of the sample statistic is obtained using Excel formula. From EXCEL formula “=CHISQ.DIST.RT(11.5, 4)”, the P-value is 0.021. Decision rule: 1. If the P-value is less that the significance level, reject the null hypothesis. 2. Otherwise fail to reject the null hypothesis. Conclusion: Here, the P-value is less than the significance level (=5%). Therefore, by the decision rule, Reject the null hypothesis. Question #18 You are conducting a multinomial Chi-Square Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies: 𝐻0 : 𝑝𝐴 = 0.3, 𝑝𝐵 = 0.1; 𝑝𝐶 = 0.1; 𝑝𝐷 = 0.5 Complete the table. Report all answers accurate to three decimal places. Category A B ( C D Observed Frequency 24 12 7 44 Expected Frequency What is the chi-square test-statistic for this data? ) 𝜒2 = What is the P-Value? P-Value= Answer: The expected values are, Category A B ( C D Observed Frequency 24 12 7 44 Proportion 0.3 0.1 0.1 0.5 Expected Frequency 87  0.3 = 26.1 87  0.1 = 8.7 ) 87  0.1 = 8.7 87  0.5 = 43.5 The chi-square test-statistic is, 𝜒2 = ∑ = (𝑂 − 𝐸)2 𝐸 (24 − 26.12 (12 − 8.7) 2 (7 − 8.7) 2 (44 − 43.5) 2 + + + 26.1 8.7 8.7 43.5 =0.169+1.252+0.332+0.006 =1.759 Thus, the chi-square test-statistic is 1.759. The degrees of freedom is, df=k-1 =4-1 =3 The degrees of freedom is 3. Computation of P-value: The P-value of 𝜒 2 -distribution for test statistic 1.759 at 3 degrees of freedom can be obtained using the excel formula “=CHISQ.DIST.RT(1.759,3)”. The P-value is 0.624. Thus, the P-value is 0.624. Question #19 You are conducting a multinomial hypothesis test (𝛼 = 0.05) for the claim that all 5 categories are equally likely to be selected. Category A B ( C D E Observed Frequency 25 13 ) 21 8 9 What is the chi-square test-statistic for this data? 𝜒2 = Answer: Given: ∝= 0.05 . k=5 From the provided data we have the computations as, Category A B C ( D E Total Observed Frequency (Oi ) 25 13 21 8 9 76 Expected Frequency ( Ei ) Oi − Ei 76 / 5 = 15.2 9.8 76 / 5 = 15.2 −2.2 76 / 5 = 15.2 5.8 76 / 5 = 15.2 −7.2 76 / 5 = 15.2 −6.2 76 0 (Oi − Ei ) 2 / Ei 6.318421 0.318421 2.213158 ) 3.410526 2.528947 14.78947 Thus, the chi-square test statistic for the given data is 14.78947. The degrees of freedom for this test is k -1 = 5-1 = 4 The p-value for this sample is given by using excel as below: The p-value is 0.0052. Question #20 You are conducting a multinomial hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Category A B ( C D E Observed Frequency 18 10 15 17 17 Expected Frequency Squared Pearson Residual ) What is the chi-square test-statistic for this data? 𝜒2 = Report all answers accurate to three decimal places. Answer: The given observed frequencies are as follows: Category ( Observed Frequency A 18 B 10 C 15 D 17 E 17 ) Calculations: The expected frequencies are as follows: Expected Frequency = ∑ Observed Frequency 𝑛 = 77 5 = 15.4 Completing the table: ( Category Observed Frequency Expected Frequency A 18 15.4 B 10 15.4 C 15 15.4 D 17 15.4 E 17 15.4 Squared Pearson Residual (18 − 15.4) 2 15.4 (10 − 15.4) 2 15.4 (15 − 15.4) 2 15.4 (17 − 15.4) 2 15.4 (17 − 15.4) 2 15.4 (Oi − Ei ) Ei = 0.4390 = 1.8935 = 0.0104 = 0.1662 = 0.1662 )