One Friday night, there were 42 carry-out orders at Ashoka Curry Express. 41. 49 43.57 21.23 27.41 42. 18 44.00 38.53 19.28 20.64 40.89 46.87 19.82 22.74 18.69 42.64 28.75 17.85 22. 01 44.79 31. 11 31.87 47.00 42. 15 30.94 25. 14 26.22 12. 98 36.50 18.46 17.86 13.09 30.96 40. 49 13.80 45. 28 44.80 15. 62 29.02 42. 91 29.20 26. 66 16. 16 Click here for the Excel Data File (a) Estimate the mean and standard deviation from the sample. (Round your answers to 4 decimal places.) Sample mean Sample standard deviation (b-1) Do the chi-square test at a = .025 (define bins by using method 3 (equal expected frequencies). Use 8 bins). (Perform a normal goodness-of-fit test for a = .025. Round your answers to 4 decimal places. Do not round your intermediate calculations.) Chi-square degrees of freedom p-valueOne Friday night, there were 42 carry-out orders at Ashoka Curry Express. 41.49 43.57 21.23 27.41 42. 18 44.00 38.53 19.28 20.64 40.89 46.87 19.82 22.74 18.69 42.64 28.75 17.85 22. 01 44.79 31. 11 31.87 47.00 42. 15 30.94 25.14 26.22 12.98 36.50 18.46 17.86 13.09 30.96 40.49 13.80 45.28 44.80 15.62 29.02 42.91 29.20 26.66 16.16 Click here for the Excel Data File (a) Estimate the mean and standard deviation from the sample. (Round your answers to 4 decimal places.) Answer is complete and correct. 30.2762 Sample mean Sample standard deviation 11.1284 (b-1) Do the chi-square test at a = .025 (define bins by using method 3 (equal expected frequencies). Use 8 bins). (Perform a normal goodness-of-fit test for a = .025. Round your answers to 4 decimal places. Do not round your intermediate calculations.) * Answer is complete but not entirely correct. Chi-square O degrees of freedom p-valueOne Friday night, there were 42 carry-out orders at Ashoka Curry Express. 41.49 43.57 21.23 27.41 42. 18 44.00 38.53 19.28 20.64 40.89 46.87 19. 82 22.74 18.69 42.64 28.75 17.85 22.01 44. 79 31. 11 31.87 47.00 42. 15 30.94 25. 14 26.22 12.98 36.50 18.46 17.86 13.09 30.96 40. 49 13.80 45.28 44.80 15.62 29.02 42.91 29.20 26.66 16. 16 Click here for the Excel Data File (a) Estimate the mean and standard deviation from the sample. (Round your answers to 4 decimal places.) Answer is complete and correct. Sample mean 30.2762 Sample standard deviation 11.1284 (b-1) Do the chi-square test at a = .025 (define bins by using method 3 (equal expected frequencies). Use 8 bins). (Perform a normal goodness-of-fit test for a = .025. Round your answers to 4 decimal places. Do not round your intermediate calculations.) Answer is complete but not entirely correct. Chi-square 12.0952 degrees of freedom 51 p-value 0.0975 XED V 0 httpszllprod.readerui.prod1mheducation.cornlepub/smSSBeb/data-uuid-79b1ad9243 (L (D [I] 33 m Question 4 - Chapter 15 Homework Connect 0 15.5 Normal Chi-Square Goodness-of-Fit Test :2 Q A3 a d For a chi-square GOF test, degrees of freedom are (if = c - m - l, where c is the number of classes used in the test and m is the number of parameters estimated. Because two parameters, [1 and a, are estimated from the sample of Kiss weights, we use m = 2. We need at least four bins to ensure at least 1 degree of freedom, while Cochran's Rule (at least 5 expected observations per bin) suggests a maximum of 7 bins for n = 35 data points (because 35/7 = 5). It therefore seems reasonable to use 6 bins for our GOF test with (if = c - m - l = 6 - 2 - 1 = 3. The GOF test in Table 15.20 shows that the chi-square test statistic (3.571) is not signicant at a = .10 (1210 = 6.251). The pvalue =CHISQ.DIST.RT(3.571,3) = .312 indicates that such a result would be expected about 312 times in 1,000 samples if the population were normal. Bin ve (highlighted) contributes heavily to the chi-square statistic. Table 15.20 Chi-Square Test for Normality of 35 Kiss Weights x f,- e]- Under 4.688 6 5.8333 4.688