Advanced Math Topics

Finals Review: Chapters 12 & 13

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.

Yes 28 36 11

No 22 44 9

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.

3.0-4.0 G.P.A.

2.0-2.99 G.P.A.

Below 2.0 G.P.A.

Yes 28 36 11

No 22 44 9

Total: 50 80 20

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 columnmust add to the total E = 25 E = 40 E = 10

Χ2 = Σ (O – E)2

E = (28 – 25)2

25 = 0.36

Do this for all six boxes and sum them up.

Χ2 = 0.36 + 0.40 + 0.10 + 0.36 + 0.40 + 0.10 = 1.72Find the critical value using d.f. = 3-1 = 2 and look under the 0.05 column

Χ2 = 5.991

Since our test statistic (1.72) is less than the critical value (5.991), we acceptthe null hypothesis. There is not a significant difference between the responsesof students with different GPA’s.

The null hypothesis is that the proportionsof yes/no responses is not different betweenthe columns.

1) The number of phone calls received per day by a local company is as follows:

M T W TH F

173 153 146 182 193

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 E = 169.4 E = 169.4 E = 169.4 E = 169.4

Χ2 = Σ (O – E)2 E

=(173 – 169.4)2

169.4 +

9.1216Find the critical value using d.f. = 5-1 = 4 and look under the 0.05 column

Χ2 = 9.488

Since our test statistic (9.1215) is less than the critical value (9.488), we acceptthe 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 =

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.

Color E O

Black 75 71

Gray 60 69

White 45 40

Χ2 = 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

1) Is eye color independent of hair color? Use a 5% level of significance.

Brown Eyes

Blue Eyes

Light Hair 10 33

Dark Hair 44 13

Total

Total

43

57

4654 100

E = 43(54)/100E = 23.22

E = 43(46)/100E = 19.78

E = 57(54)/100E = 30.78

E = 57(46)/100E = 26.22

Χ2 = Σ (O – E)2

E = (10 – 23.22)2

23.22 = 7.527

Do this for all four boxes and sum them up.

Χ2 = 7.527 + 8.836 + 5.678 + 6.665 = 28.706The degrees of freedom is (# of rows – 1)(# of columns – 1).

(2 – 1)(2 – 1) = 1

3.841vs.

Since our test statistic is larger,eye color is dependent on hair color.

4) A study wants to compare the cost of hospitals across the country. The prices showthe 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 $427 382 502 476

South 391 402 427 501

East 517 378 476 409

West 501 499 404 428

To solve this type of question, use an ANOVA table.

An ANOVA Table

(1)

(4)

(7)

(9)

(2)

(5)

(8)

(3)

(6)

Factor of theExperiment

Error

Total

Sum of Squares

Degreesof freedom

Mean Square f-ratio

Σ(row total2)# of columns

- ___Total2__ # of Boxes

(sum of each #2) - Σ(row total2)# of columns

Add cells 1 & 2

r – 1r = # ofrows

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 fromCell 9.

North $427 382 502 476

South 391 402 427 501

East 517 378 476 409

West 501 499 404 428

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 revealthe answer.

Bring in your textbooks tomorrow and sign up for Schoolloop for homework.