1. Advanced Math Topics 12.3 Goodness of Fit

2. The chi-square statistic is used to determine if two or more sample percentages are significantly different. It can also be used to determine whether an observed frequency distribution is in agreement with mathematical expectation. You roll a die 120 times and get the following results. FaceObserved Frequency 118 221 317 421 519 624 Do the differences seem reasonable (purely to chance) or are they significant? Is the die honest? We can find the expected value and use the chi-square statistic to answer these questions. We will test the goodness of fit, that is, to determine whether the observed frequencies fit with what was expected.

3. 1) The number of phone calls received per day by a local company is as follows: MTWTHF 173153146182193 Using a 5% level of significance, test the null hypothesis that the number of calls received is independent of the day of the week. We first calculate the number of expected calls per day. If the number of calls per day is independent of the day of the week, we would expect to receive… 173 + 153 + 146 + 182 + 193 5 =169.4 E = 169.4 Χ 2 = Σ (O – E) 2 E = (173 – 169.4) 2 169.4 + 9.1216 Find the critical value using d.f. = 5-1 = 4 and look under the 0.05 column Χ 2 = 9.488 Since our test statistic (9.1215) is less than the critical value (9.488), we accept the null hypothesis. The phone calls received are independent of the day of the week. (153 – 169.4) 2 169.4 + (146 – 169.4) 2 169.4 + (182 – 169.4) 2 169.4 + (193 – 169.4) 2 169.4 =

4. 2) A scientist claims that when two rats mate, the offspring will be black, gray, and white in the proportion 5:4:3. Out of 180 newborn rats, 71 are black, 69 are gray, and 40 are white. Can we accept the scientist’s claim? Use a 5% level of significance. 5 + 4 + 3 = 12 The expected probability of a black rat is 5/12. The expected probability of a gray rat is 4/12. The expected probability of a white rat is 3/12. The expected frequency of a black rat is 180(5/12) = 75. The expected frequency of a gray rat is 180(4/12) = 60. The expected frequency of a white rat is 180(3/12) = 45. ColorEO Black7571 Gray6069 White4540 Χ 2 = Σ (O – E) 2 E = (71 – 75) 2 75 + 2.1189 We accept the scientist’s claim! (69 – 60) 2 60 + (40 – 45) 2 45 = vs. 5.991

5. From the HW:P. 613 2) There are five toll booths at the entrance to a tunnel. An toll booth worker notices that 800 randomly selected cars use the toll booths in the following frequencies. Using a 5% level of significance, test the null hypothesis that each toll booth is used equally. Booth ## of Cars 1130 2194 3196 4106 5174 Answer: 40.4 vs. 9.488. Reject the null hypothesis that the booths are used equally.

6. From the HW:P. 613 5) A company claims that a multi-vitamin product will contain vitamins A, B, C, and D in the ratio of 9:7:4:3. A randomly selected bottle had 261 A, 213 B, 111 C, and 105 D vitamins. Using a 1% level of significance, test the null hypothesis that the frequencies are in the claimed ratio. Answer: 3.5179 vs. 11.345. Accept the null hypothesis that the vitamins are in the claimed ratio.

7. HW P. 613 #2-7