Question

Questions 45-46: Consider the following partial table of a sample of 60 M&Ms. We count how many of each color are present. We are testing the null hypothesis that each color has the stated proportion. Observed Counts Assumed Proportions Expected Counts 0.24 13 14 0.20 10 0.16 Color Blue Orange Green Yellow Red Brown Total 11 0.14 6 0.13 6 0.13 60 1.00 60 45. If the null hypothesis is correct, how many orange M&Ms do we expect to see in our sample?:* Oa. 8.4 Ob. 9.6 Oc. 12 Od. 14.4 46. Find the p-value.:* O a. 0.0381 Ob. 0.0873 Oc. 0.1679 O d. 0.8321

Answer 1

Blue Answer consider the table Color Observed counts (0) Expected counts (E;) (Oi-E:)?! Ei 13 0.24 X 60 = 14.4 0.1361 Orange 14 0.20x60= 12.0 0,3333 Green 10 0.16X 60 = 9.6 0,0167 Yellow 0.14460=0.4 0.0048 Red 6 0.13X60 = 7.8 0.4154 Brown 6 0.13 X 605 7.8 0.41 54 Total 60 60 2. 1217 (455) If the null hypothesis is correct, the expected no el orange Mema is 0.20 X 60 = 12. Hence (C) is correct (46) x2 = [ (0;-£;)? = 2.12.17 Then p-value in Ei P-value = Pr [x'>x?] where 82 follows Chi-square distribution with n=5 = Pr[x?>2.1217) degrees of freedom 0.8321 Hence option (d) is correct

Are (47) Asuming that the null hypothesis it true in Chi-square incependence test, the expected number of females preferring. Twilight is 28X108 19.76 153 Hence option (6) option (6) is correct. 103