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CHI-SQUARE

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2

Learning Guide IV – Chapter 11

Chi-square

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2

“GOODNESS-OF-FIT”

Observed frequencies in 120 tosses of a die are listed in the table below. At the 5% level, is there any reason

to believe that this die is “loaded?” State the probabilities and calculate the expected values.

1.

2.

3. H0: n categories =

H1: df =

4.

5. Decision:

6. Conclusion:

…………………………………………………………………………………………………………………………………………………………..

7. A marketing firm randomly selects 500 music listeners from a broadcast region and asks their

preference in music. The observed results are shown in the table. Find the expected frequencies.

……………………………………………………………………………………………………………………………………

8. A television producer reports that the audience for his program will be distributed in the table. In a

sample of 250 participants, the observed counts are shown. Find the expected numbers.

9. What is the chi-square statistic 10. Is the difference of the “actual” from

for these data? the “expected” significant?

A. 17.61 B. 12.97 C. Yes D. No

…………………………………………………………………………………………………………………………………..

11. A chi-square test at the 5% significance level is performed of a data set with 8 categories.

What is the critical value of this test?

A. 12.592 B. 14.067 C. 15.507 D. 16.919

# on die 1 2 3 4 5 6 TOTALS

Probability, P(x)

Observed frequency 17 25 17 23 15 23 120

Expected frequency

( )

2

2

Statistic

observed exp ected

exp ected

−

=

Type of Music Classical Country Gospel Oldies Pop Rock TOTALS

Probability, P(x) 4% 36% 11% 2% 18% 29% 100%

Observed 8 210 72 10 75 125 500

Expected

Age group Under 25 25-44 45-65 Over 65 TOTALS

Probability distribution 0.14 0.22 0.40 0.24 1.00

Actual number 35 30 125 60 250

Expected number

Critical value =

( )

2 .05

=

α

=

1

Label the critical value(s) and test

statistic. Shade the rejection region(s).

It is believed that the proportions of people with A, B, AB, and O blood types in the population are 0.4, 0.2, 0.1,

and 0.3, respectfully. When 400 randomly selected people were examined, the observed numbers are

recorded below. Calculate the expected numbers. At the 5% level, do these data support the stated belief?

At the 1% level, do these data support the stated belief?

12.

13.

14. H0: n categories =

H1: df =

15.

16. Conclusion at the 5% level:

17. Conclusion at the 1% level:

…………………………………………………………………………………………………………………………………………………………..

According to the Mendelian law of segregation in genetics, when two doubly heterozygous walnut comb fowls

are crossed, the offspring consist of four types of fowls: walnut, rose, pea and single. The expected

frequencies are in the ratio 9:3:3:1. Among 320 offspring, it was found that there were 160 walnut, 70 rose, 65

pea and 25 single. Are these probabilities for each type in keeping with the Mendelian Law at

= 0.05

?

18.

19.

20. H0: n categories =

H1: df =

21.

22. Decision:

23. Conclusion:

Blood type A B AB O TOTALS

Probability, P(x) 0.4 0.2 0.1 0.3 1.0

Observed frequency 148 96 50 106 400

Expected frequency

( )

2

2

Statistic

observed exp ected

exp ected

−

=

Fowl types Walnut Rose Pea Single TOTALS

Probability, P(x)

9 16 16 16

Observed frequency 160 70 65 25 320

Expected frequency

( )

2

2

Statistic

observed exp ected

exp ected

−

=

Critical value =

( )

2

.05

=

Critical value =

( )

2

.01

=

=

Critical value =

( )

2

.05

=

2

Label the critical value(s) and test

statistic. Shade the rejection region(s).

Label the critical value(s) and test

statistic. Shade the rejection region(s).

Suppose, according to a survey conducted in 1960, the probability age distribution of an adult over 20 years of

age is as given in the table below. In 1993, a sample of 1000 adults was interviewed and the frequency

distribution was recorded in row 3. Using a 5% level, test the hypotheses that the probability distribution in

1993 was the same as that in 1960.

24.

25.

26. H0: n categories =

H1: df =

27.

28. Decision:

29. Conclusion:

…………………………………………………………………………………………………………………………………………………………..

Chi-square

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2

“CONTINGENCY TABLES”

30. To perform a Chi-square test of independence on a table with 5 rows and 3 columns, how many

degrees of freedom are used? Note: Degrees of Freedom = (Row – 1)(Column – 1)

A. 8 B. 12 C. 13 D. 15

………………………………………………………………………………………………………………………………………………..

31. What is the critical value at the 5% level of significance?

E. 14.067 F. 15.507 G. 23.685 H. 24.996

…………………………………………………………………………………………………………………………………………………………..

32. The Chi-square statistic for a test of independence is 17.86. If the table has 4 rows and 2 columns,

what is the Chi-square critical value at the 1% significance level?

A. 11.345 B. 20.090 C. 6.251 D. 7.815

………………………………………………………………………………………………………………………………………………..

33. What decision is made regarding

H0

, the hypothesis of independence?

E. Independence is rejected F. Independence is not rejected

…………………………………………………………………………………………………………………………………

34. A calculator displays the following:

2

−

TEST

2

3.493357223

p .321624691

df 3

=

=

=

25

Age group 20-29 30-39 40-49 50-59 60-69 70+ TOTALS

Probability, P(x) 0.25 0.23 0.20 0.15 0.08 0.09 1.00

Observed (1993) 270 202 180 160 88 100 1000

Expected (1960)

( )

2

2

Statistic

observed exp ected

exp ected

−

=

Critical value =

( )

2

.05

=

Using a 5% level, you decide

A. the test statistic is not in the rejection region

B. the rejection region is

2

7.815

C. the p-value is greater than alpha

D. the null hypothesis H0

should not be rejected

E. all of these are true 3

Label the critical value(s) and test

statistic. Shade the rejection region(s).

Five hundred adults participated in a comparison of the effectiveness of 3 arthritic pain relievers. Each

participant used 1 of the 3 medications for 1 month and then asked if the product was effective. Perform a test

of independence at the 1% level of significance. Is the type of pain reliever independent of effectiveness?

35. H0: df =

H1:

α

=

36.

37. Record the expected values in the table below from your calculator.

38. Decision:

39. Conclusion:

…………………………………………………………………………………………………………………………………

A particular city can watch national news on affiliate stations of ABC, NBC, CBS, and PBS. A researcher

wishes to know whether there is any relationship between political philosophy (conservative, moderate, or

liberal) and preferred news program among those who regularly watch the national news. A random sample of

300 regular news watchers was selected. At the 5% level of significance, can you reject the null hypothesis

that party affiliation and national news stations are independent on the basis of the

2

Statistic

?

40. H0: df =

H1:

α

=

Continued on next page

Pain Reliever

Effective A B C TOTALS

YES 115 78 140 333

NO 60 72 35 167

TOTALS 175 150 175 500

Pain Reliever

Effective A B C TOTALS

YES 333

NO 167

TOTALS 175 150 175 500

ABC CBS NBC PBS TOTALS

Liberal 20 20 25 15 80

Moderate 45 35 50 20 150

Conservative 15 40 10 5 70

TOTALS 80 95 85 40 300

Critical value =

( )

2

.01

=

Calculated

2

= Statistic

p-value =

4

Label the critical value(s) and test

statistic. Shade the rejection

region(s).

41.

42. Record the expected values in the table below from your calculator.

43. Decision:

44. Conclusion:

…………………………………………………………………………………………………………………………………………………………..

An insurance company is studying the relationship between types of crashes and the vehicle involved. 3207

vehicle crashes are randomly selected and the data organized in the contingency table. Using a 5% level of

significance, can the company conclude that the type of crash depends on the type of vehicle involved?

45. H0: df =

H1:

α

=

46.

47. Record the expected values in the table below from your calculator.

48. Decision:

49. Conclusion:

ABC CBS NBC PBS TOTALS

Liberal 80

Moderate 150

Conservative 70

TOTALS 80 95 85 40 300

Car Truck/SUV Cargo/Large Van TOTALS

Single-vehicle 895 493 45 1433

Multiple-vehicle 1400 336 38 1774

TOTALS 2295 829 83 3207

Car Truck/SUV Cargo/Large Van TOTALS

Single-vehicle 1433

Multiple-vehicle 1774

TOTALS 2295 829 83 3207

Critical value =

( )

2

.05

=

Calculated

2

= Statistic

p-value =

Critical value =

( )

2

.05

=

Calculated

2

= Statistic

p-value =

5

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