Question

PART 1:

To target the right age-group of people, a marketing consultant must find which age-group purchases from home-shopping channels on TVs more frequently. According to management of TeleSell24/7, a home-shopping store on TV, about 30% of the online-music-downloaders are in their fifties, but the marketing consultant does not believe in that figure. To test this he selects a random sample of 256 online-music-downloaders and finds 64 of them are in their fifties.

1. The value of the test-statistic is: Answer to 2 decimal places.

PART 2:

The following two-way table of counts summarizes whether respondents smoked or not and whether they have had ever divorced or not for persons who had ever been married.

	Ever Divorced?
Smoke?	Yes	No
Yes	210	278
No	440	532

1. Among those who smoked, what percentage has ever been divorced? [Answer to 2 decimal places. Do not type % symbol in the box.]  %

2. Among those who has ever been divorced, what percentage smoked? [Answer to 2 decimal places. Do not type % symbol in the box.]  %

Next we intend to test if smoking habits and being divorced are related or not.
3. What is the expected frequency of smoker and ever being divorced? [Answer to 2 decimal places.]

4. What is the expected frequency of smoker and never being divorced? [Answer to 2 decimal places.]

5. What is the expected frequency of non-smoker and ever being divorced? [Answer to 2 decimal places.]

6. What is the expected frequency of non-smoker and never being divorced? [Answer to 2 decimal places.]

7. To test independence between smoking habits and being divorced, what is the value of chi-square test statistic? [Answer to 3 decimal places.]

Suppose we are testing:

Null hypothesis: smoking habit and ever being divorced are not related,
against
Alternative hypothesis: smoking habit and ever being divorced are related.

8. If the p-value associated to the ch-square test-statistics is 0.418 and the level of significance is 5%, what will be your conclusion?
Not enough information to reach a decision
Do not reject null hypothesis
Reject null hypothesis

Answer 1

Part 1:

n = 256, x = 64

p̄ = x/n = 0.25

Test statistic:

z =(p̄ -p)/(√(p*(1-p)/n)) = -1.75

---------------------------

Part 2:

	Divorced Yes	No	Total
Smoke Yes	210	278	488
No	440	532	972
Total	650	810	1460

1. Among those who smoked, percentage that has ever been divorced = 210/488 = 43.03%

2. Among those who has ever been divorced, percentage that smoked = 210/650 = 32.31%

3. Expected frequency of smoker and ever being divorced = 650 * 488 / 1460 = 217.26

4. Expected frequency of smoker and never being divorced = 810 * 488 / 1460 = 270.74

5. Expected frequency of non-smoker and ever being divorced = 650 * 972 / 1460 = 432.74

6. Expected frequency of non-smoker and never being divorced = 810 * 972 / 1460 = 539.26

7.

	(fo-fe)²/fe
Smoke Yes	(210 - 217.26)²/217.26 = 0.2426	(278 - 270.74)²/270.74 = 0.1947
No	(440 - 432.74)²/432.74 = 0.1218	(532 - 539.26)²/539.26 = 0.0977

Chi-square test statistic:

χ² = ∑ ((fo-fe)²/fe) = 0.6569

Null hypothesis: smoking habit and ever being divorced are not related.

Alternative hypothesis: smoking habit and ever being divorced are related.

8. As p-value = 0.418 > 0.05, we Do not reject the null hypothesis.