1

Always be mindful of the kindness and not the faults of others.

Categorical Data

Sections 10.1 to 10.5Estimation for proportionsTests for proportionsChi-square tests

3

Example

Researchers in the development of new treatments for cancer patients often evaluate the effectiveness of new therapies by reporting the proportion of patients who survive for a specified period of time after completion of the treatment. A new treatment of 870 patients with lung cancer resulted in 330 survived at least 5 years.

4

Example

Estimate , the proportion of all patients with lung cancer who would survive at least 5 years after being administered this treatment

How much would you estimate the proportion as?

5

Distribution of Sample Proportion

Y: the number of successes in the n trials (independent and identical trials)

What’s the distribution of Y?

Sample proportion,

n

Y

n

)1(ˆ

ˆ

6

Distribution of Sample Proportion

When n≥ 5 and n(1-≥ 5, the distribution of Y can be approximated by a normal distribution.

(approximate) (1-) Confidence Interval for :

Optional: (exact) C.I. for for small sample

ˆ2/ ˆˆ Z

7

Sample Size

2

22/ )1(

E

Zn

Where E is the largest tolerable error at (1- confidence level.

8

Test for a Large Sample

When n≥ 5 and n(1-≥ 5, the test statistic is:

n

Z)1(

ˆ

00

0

9

Inference about 2 Proportions

Notation:

Population 1 Population 2

Proportion

Sample size n1 n2

# of successes y1 y2

Sample proportion

1

11ˆ n

y

2

22ˆ

n

y

10

Estimation for

Point estimate: 21 ˆˆ

21ˆˆ 21

1

22

1

11ˆˆ

)1()1(11 nn

11

Estimation for

(1-) Confidence Interval for two large samples:

1

22

1

11ˆˆ

)ˆ1(ˆ)ˆ1(ˆˆ

21 nn

21 ˆˆ2/21 ˆˆˆ z

12

Example 10.6

A company markets a new product in the Grand Rapids and Wichita.

In Grand Rapids, the company’s advertising is based entirely on TV commercials.

In Wichita, based on a balanced mix of TV, radio, newspaper, and magazine.

2 months after the ad campaign begins, the company conducts surveys to determine consumer awareness of the product.

13

Example 10.6: Data Set

Grand Rapids Wichita

# of interviewed 608 527

# of aware 392 413

Q: Calculate a 95% C.I. for the regional difference in the proportion of all consumers who are aware of the product.

14

Example 10.6 (conti.)

Conduct a test at =0.05 to verify if there are >10% more Wichita consumers than Grand Rapids consumers aware of the product.

15

Test for Large Samples)

When n1≥ 5 and n1(1-≥ 5; n2≥ 5 and n2(1-≥ 5, the test statistic of Ho: p1-p2=d is

Optional: Fisher Exact Test (p.511)

1

22

1

11

21

)ˆ1(ˆ)ˆ1(ˆ

)ˆˆ(

nn

dZ

Minitab

Z procedure for one proportion:

Stat >> Basic Statistics >>1 proportion Z procedure for two proportions:

Stat >> Basic Statistics >>2 proportion Sample size calculation:

Stat >>Power & Sample size>>1 proportion or 2 proportion

Stat >>Power & Sample size>>sample size for estimation

16

17

Chi-Square Goodness of Fit Test

More than two possible outcomes per trial the multinomial experiment

1. The experiment consists of n identical trials.

2. Each trial results in one of k outcomes with probabilities ...k.

Y=(Y1,…,Yk); Yi = the # of outcome i.

18

Chi-square Goodness of Fit Test

Goal: We are interested in testing a hypothesized distribution of Y (i.e. a set of i’s values).

Hypotheses:

Ho: i = io for all i vs. Ha: Ho is false

19

Chi-square Goodness of Fit Test

Test Statistic:

ni = the observed YiEi = the expected Yi = nio

i i

ii

E

En 22 )(

20

Chi-square Goodness of Fit Test

Rejection Region:

Reject Ho if where df=k-1.

Note:

This test can be trusted only when 80% of more cells of the Ei’s are at least 5.

2,

2df

21

Example 10.10

Category Hypothesized % Observed counts

Marked decrease 50 120

Moderate decrease 25 60

Slight decrease 10 10

Stationary of slight increase

15 10

Minitab: Stat >> Tables >> Chi-Square Goodness-of-Fit Test(One Variable)

22

Example 10.11

23

Contingency Table(Example 10.12)

nijAge Category

Severity of skin disease

1 2 3 4 Total ni*

1 15 32 18 5 70

2 8 29 23 18 78

3 1 20 25 22 68

Total n*j 24 81 66 45 216 = n

24

Contingency Table

2 categorical variables:

row and column indexed by i and j, respectively

If they are independent, then

n

nnE ji

ijˆ

25

Test for Independence of 2 Var’s

Hypotheses:

Ho: the row and column variables are independent

Ha: they are dependent

Test Statistic:

ji ij

ijij

E

En

,

22

ˆ)ˆ(

26

Test for Independence of 2 Var’s

Rejection Region:

Reject Ho if where df=(r-1)(c-1).

Note:

This test can be trusted only when 80% of more cells of the are at least 5.

2,

2df

ijE

Minitab: Stat >> Tables >> Cross Tabulation and Chi-square

Tabulated Statistics: C1, Worksheet columns

Rows: C1 Columns: Worksheet columns

1 2 3 4 5 All

A 32 87 91 46 44 300

3.56 9.67 10.11 5.11 4.89 33.33

42 107 78 34 39

B 53 141 76 20 10 300

5.89 15.67 8.44 2.22 1.11 33.33

42 107 78 34 39

C 41 93 67 36 63 300

4.56 10.33 7.44 4.00 7.00 33.33

42 107 78 34 39

All 126 321 234 102 117 900

14.00 35.67 26.00 11.33 13.00 100.00

Cell Contents: Count

% of Total

Expected count

Pearson Chi-Square = 72.521, DF = 8, P-Value = 0.000

Likelihood Ratio Chi-Square = 79.263, DF = 8, P-Value = 0.000

27

Example 10.12