Question

You supervise two outpatient facilities and want to compare the patient satisfaction at each. You decide that a score of 8.5, or better, out of 10 will be considered "good satisfaction", and a score of less than 8.5out of 10, will be considered "poor satisfaction". You conduct a survey of patients at each facility over the same three month interval and gather the following data: Total Facility #1 Facility #2 Total #with average score 8.5 or higher 437 353 790 #with average score 8.4 or lower 154 164 318 591 517 1,108 You perform a chi square test to determine whether the differences in scores are independent of the facility from which they came. You will have to calculate the expected values for each of the cells (lecture 14, slides 4,5). It would be nice if Excel could calculate the expected values from the actual data ... but it doesn't . In Excel, create two 2x2 tables; one with the actual values and one with the expected values. Use the CHISQ.TEST function to calculate the chi square value for the data. When selecting the “Actual range and the "Expected range" for your data entry, select the 2x2 values (not including the totals) for *both* of the facilities for the actual range and select the 2x2 values (not including the totals) for both of the facilities for the expected range. E.g., for the actual range, you'd select the highlighted cells: Total Facility #1 Facility #2 #with average score 8.5 or higher 437 353 790 #with average score 8.4 or lower 154 164 318 591 517 1.108 Total Similarly, for the expected range (selected from the table of expected values that you created). The result that Excel displays when you run the chi square test is the p-value. You don't have to do anything else a. Using Excel, determine whether the results are independent of the facility from which they came. Copy and paste your Excel data and result. (Just copy and paste the portion of the Excel page that has your work on it; do not copy and paste the entire Excel page .... they're huge!) (10) b. What is your conclusion about patient satisfaction at each of the facilities? (5)

Answer 1

a.

Given table data is as below MATRIX col1 col2 TOTALS row 1 437 154 591 row 2 353 164 517 TOTALS 790 318 N = 1108 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N ------------------------------------------------------------------ expected frequencies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 421.381 169.619 row 2 368.619 148.381 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 437 421.381 15.619 243.95 0.579 154 169.619 -15.619 243.95 1.438 353 368.619 -15.619 243.95 0.662 164 148.381 15.619 243.95 1.644 ᴪ^2 o = 4.323 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, α = 0.05 from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =3.841 since our test is right tailed,reject Ho when ᴪ^2 o > 3.841 we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei from the table , ᴪ^2 o = 4.323 critical value the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841 we got | ᴪ^2| =4.323 & | ᴪ^2 α | =3.841 make decision hence value of | ᴪ^2 o | > | ᴪ^2 α| and here we reject Ho ᴪ^2 p_value =0.038 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent test statistic: 4.323 critical value: 3.841 p-value:0.038 decision: reject Ho

b.

we have enough evidence to support the claim that whether the differences in score of independent of the facility from whch they came.