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A study reported that 47% of 16- to 29-year olds, 41% of 30- to 49-year-olds, an

ID: 3175547 • Letter: A

Question

A study reported that 47% of 16- to 29-year olds, 41% of 30- to 49-year-olds, and 34% of 50- to 64-year-olds often listened to rock music. Suppose that the study was based on a sample size of 200 in each group. Complete parts (a) through (c).

The test statistic is X2STAT=____________________

(Round to three decimal places as needed.)

The critical value for =0.05 is ______________________

(Round to three decimal places as needed.)

(1) ____________________H0. There (2)_____________________evidence of a difference among the age groups with respect to the proportion who often listened to rock music.

b. Determine the p-value in (a) and interpret its meaning.

The p-value is__________________

(Round to three decimal places as needed.)

What does the p-value mean?

A. The p-value is the probability of rejecting the null hypothesis if the test were repeated for different samples.

B. The p-value is the probability of obtaining a result at least as extreme as the one obtained with this sample given that the null hypothesis is true.

C. The p-value is the probability that the null hypothesis is true.

D. The p-value is the probability of falsely rejecting the null hypothesis.

c. If appropriate, use the Marascuilo procedure and =0.05 to determine which age groups are different.

Select all that apply.

A.30-to 49-year-olds and 50- to 64-year-olds

B. 16-to 29-year-olds and 50- to 64-year-olds

C. 16- to 29-year-olds and 30- to 49-year-olds

D. The procedure is not appropriate, so no two groups are different.

(1) Reject, Do not reject

(2) is insufficient, is sufficient

Explanation / Answer

X2STAT = S [ (Oi – Ei)^2 / Ei]

Oi = observed value

Ei = expected value

200 * 0.47 = 94 and 200 – 94 = 106

200*0.41 = 82 and 200 – 82 = 118

200*0.34 = 68 and 200 – 68 = 132

From the given values we get the contingency table as:

16 to 19 year

30 to 49 year

30 to 49 year

Total

Rock

94

82

68

244

No Rock

106

118

132

356

Total

200

200

200

600

(200*244)/600 = 81.3333

(200 *356)/600 = 118.6667

(200*244)/600 = 81.3333

(200 *356)/600 = 118.6667

(200*244)/600 = 81.3333

(200 *356)/600 = 118.6667

Ei:

16 to 19 year

30 to 49 year

30 to 49 year

Total

Rock

81.3333

81.3333

81.3333

244

No Rock

118.6667

118.6667

118.6667

356

Total

200

200

200

600

Oi – Ei:

16 to 19 year

30 to 49 year

30 to 49 year

Rock

12.6667

0.6667

-13.3333

No Rock

-12.6667

-0.6667

13.3333

(Oi – Ei)^2:

16 to 19 year

30 to 49 year

30 to 49 year

Rock

160.4444

0.4444

177.7778

No Rock

160.4444

0.4444

177.7778

16 to 19 year

30 to 49 year

30 to 49 year

Total

Rock

1.973

0.005

2.186

4.164

No Rock

1.352

0.004

1.498

2.854

We add them all and we get 7.018

The test statistics is X2STAT = 7.018

The degrees of freedom = (3-1)*(2-1) = 2

Level of significance is .05

               The critical value for a =.05 is 5.991

               p-value = 0.030

            The p-value is .030

             B. The p-value is the probability of obtaining a result at least as extreme as the one-obtained with this sample given that the null hypothesis is true.

16 to 19 year

30 to 49 year

30 to 49 year

Total

Rock

94

82

68

244

No Rock

106

118

132

356

Total

200

200

200

600