1. Advanced Math Topics Finals Review: Chapters 12 & 13

2. A teacher surveyed his students to see if they would take a class from him again. He organized the results by G.P.A. in the table below: 3.0-4.0 G.P.A. 2.0-2.99 G.P.A. Below 2.0 G.P.A. Yes283611 No22449 Is there a significant difference between the answers for each G.P.A.? Use 5% level of significance. We can use the Chi-square statistic to answer the question.

3. 3.0-4.0 G.P.A. 2.0-2.99 G.P.A. Below 2.0 G.P.A. Yes283611 No22449 Total:508020 p = sum of the top row total sample size 75 150 =0.5= E = 0.5(50) = (25) E = 0.5(80) = (40)E = 0.5(20) = (10) The two E values in each column must add to the total E = 25E = 40E = 10 Χ 2 = Σ (O – E) 2 E = (28 – 25) 2 25 =0.36 Do this for all six boxes and sum them up. Χ 2 = 0.36 +0.40 +0.10 +0.36 +0.40 +0.10 =1.72 Find the critical value using d.f. = 3-1 = 2 and look under the 0.05 column Χ 2 = 5.991 Since our test statistic (1.72) is less than the critical value (5.991), we accept the null hypothesis. There is not a significant difference between the responses of students with different GPA’s. The null hypothesis is that the proportions of yes/no responses is not different between the columns.

4. 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 =

5. 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

6. 1) Is eye color independent of hair color? Use a 5% level of significance. Brown Eyes Blue Eyes Light Hair1033 Dark Hair4413 Total Total 43 57 4654100 E = 43(54)/100 E = 23.22 E = 43(46)/100 E = 19.78 E = 57(54)/100 E = 30.78 E = 57(46)/100 E = 26.22 Χ 2 = Σ (O – E) 2 E = (10 – 23.22) 2 23.22 =7.527 Do this for all four boxes and sum them up. Χ 2 = 7.527 +8.836 +5.678 +6.665 =28.706 The degrees of freedom is (# of rows – 1)(# of columns – 1). (2 – 1)(2 – 1) = 1 3.841 vs. Since our test statistic is larger, eye color is dependent on hair color.

7. 4) A study wants to compare the cost of hospitals across the country. The prices show the daily hospital charge from four hospitals in each region. At a 5% level, test the claim that the average daily charge is significantly different depending on the region. North $427382502476 South 391402427501 East 517378476409 West 501499404428 To solve this type of question, use an ANOVA table.

8. An ANOVA Table (1) (4) (7) (9) (2) (5) (8) (3) (6) Factor of the Experiment Error Total Sum of Squares Degrees of freedom Mean Square f-ratio Σ(row total 2 ) # of columns - ___Total 2 __ # of Boxes (sum of each # 2 ) - Σ(row total 2 ) # of columns Add cells 1 & 2 r – 1 r = # of rows r(c – 1) c = # of columns Add cells 4 & 5 Cell 1/Cell 4 Cell 2/Cell 5 Cell 7/Cell 8 Compare your test statistic to the critical value. This is found on A14-15(5%) or A16-17(1%) by looking up the d.f. for the numerator and denominator from Cell 9. North$427382502476 South391402427501 East517378476409 West501499404428

9. Prepare your green notecard(put all formulas on the card, no formulas will be given) and study Chapters 11, 12 and 13 for the Final tomorrow. A good way to study is to look at the Powerpoints from each Chapter, or the Review Powerpoints, and solve each question…then reveal the answer. Bring in your textbooks tomorrow and sign up for Schoolloop for homework.