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
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