Background: Morris Saldov conducted a study in Eastern and Central Newfoundland in 1988 to examine public attitudes towards social spending. In particular, the study tried to determine if knowing someone on public assistance (yes, no) affected one's views on social spending (too little, about right, too much). The data from the study is summarized in the table below. Too little About right Too much Total Yes No 42 7 49 15 11 26 97 16 66 25 91 Source: Morris Saldov, Public Attitudes to Social Spending in Newfoundland," Canadian Review of Social Policy, 26, November 1990, pages 10-14. Total Directions: Conduct a chi-square test for independence to determine if the association between knowing someone on public assistance and views on social spending is statistically significant. 1. Choose the correct null and alternative hypotheses. O Ho: There is no association between knowing someone on public assistance and views on social spending. Ha There is an association between knowing someone on public assistance and views on social spending. O Ho: There is an association between knowing someone on public assistance and views on social spending. Ha There is no association between knowing someone on public assistance and views on social spending. Too little 2. Compute the test statistic. Complete the following table of expected counts. (Round your answers to 3 decimal places). About right Too much Yes No Compute the value of the test statistic. (Round your answer to 2 decimal places.) x² = 3. Compute the p-value. (Round your answer to 4 decimal places.) p-value= 4. Interpret the results of the significance test. The p-value provides strong evidence against the null hypothesis. The association between knowing someone on public assistance and attitudes towards social spending is statistically significant. From a practical perspective, a person who knows someone on public assistance is 2.3 te likely to believe that too little is spent on social needs

Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a time so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts. The given data is Yes No Total Too little 42 7 49 About right 15 11 26 Too much 9 7 16 Total 66 25 91 1) We conduct a chi-square test for independence to determine if the association between knowing someone on public assistance and views on social spending is statistically significant. We know that in the chi-square test for independence, the null hypothesis represents independence (or no association) between the two categorical variables. So, the null and alternative hypotheses are H0: There is no association between knowing someone on public assistance and views on social spending. Ha: There is an association between knowing someone on public assistance and views on social spending. So, the 1st option is correct and the 2nd option is incorrect. 2) Note that the expected frequency corresponding to a cell is E=RT×CTn, where RT is the row total corresponding to that cell, CT is the column total corresponding to that cell and n is the sample size. For the first cell, RT=49 and CT=66. In this problem, n=91. So, the expected frequency corresponding to the first cell is E=RT×CTn=49×6691≈35.538 For the second cell, RT=49 and CT=25. So, the expected frequency corresponding to the second cell is E=RT×CTn=49×2591≈13.462 In this way, we compute the expected frequencies corresponding to the remaining cells. So, the table for expected counts is Yes No Too little 35.538 13.462 About right 18.857 7.143 Too much 11.604 4.396The table for the calculation of the test statistic is O E O-E2E 42 35.538 1.175 7 13.462 3.1019 15 18.857 0.7889 11 7.143 2.0827 9 11.604 0.5844 7 4.396 1.5425 Sum ---- 9.2754 From the above table, we have ∑(O-E)2E=9.2754. So, the observed value of the test statistic is χ2=∑(O-E)2E=9.2754≈9.28 The value of the test statistic is χ2=9.28.3) The p-value for the chi-square test of independence is p-value=Pχ2>χ02, where χ2 follows the chi-square distribution with df=(r-1)(c-1) and χ02 is the observed value of the test statistic. r is the number of rows and c is the number of columns in the given data. In this problem, r=3 and c=2. So, df=(r-1)(c-1)=(3-1)×(2-1)=2 Now, the p-value is p-value=Pχ2>χ02=Pχ2>9.28≈0.0097 We use excel to find Pχ2>9.28. So, p-value=0.0097.

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