Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.01 to test for a difference between the weights of discarded paper​ (in pounds) and weights of discarded plastic​ (in pounds). Household Paper Plastic   1   6.05   2.73     2   13.05   12.31     3   16.08   14.36     4   6.98   2.65     5   2.41   1.13     6   9.45   3.02     7   9.19   3.74     8   12.32   11.17     9   11.42   12.81     10   13.61   8.95     11   9.41   3.36     12   11.08   12.47     13   16.39   9.70     14   9.83   6.26     15   17.65   11.26     16   7.57   5.92     17   12.73   14.83     18   8.82   11.89     19   20.12   18.35     20   6.67   6.09     21   11.36   10.25     22   6.33   3.86     23   15.09   9.11     24   3.27   0.63     25   8.72   9.20     26   7.72   3.86     27   13.31   19.70     28   12.43   8.57     29   6.96   7.60     30   6.16   5.88     Household Paper Plastic Click the icon to view the data. In this​ example, μd is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis​ test?     A. H0​: μd=0 H1​: μd≠0   B. H0​: μd=0 H1​: μd<0   C. H0​: μd≠0 H1​: μd>0   D. H0​: μd≠0 H1​: μd=0 Identify the test statistic.   t=nothing ​(Round to two decimal places as​ needed.) Identify the​ P-value.   ​P-value=nothing ​(Round to three decimal places as​ needed.) What is the conclusion based on the hypothesis​ test?   Since the​ P-value is ▼   less greater than the significance​ level, ▼   reject fail to reject the null hypothesis. There ▼   is not is sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.

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Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of
0.01
to test for a difference between the weights of discarded paper​ (in pounds) and weights of discarded plastic​ (in pounds).
Household
Paper
Plastic
 
1
 
6.05
 
2.73
 
 
2
 
13.05
 
12.31
 
 
3
 
16.08
 
14.36
 
 
4
 
6.98
 
2.65
 
 
5
 
2.41
 
1.13
 
 
6
 
9.45
 
3.02
 
 
7
 
9.19
 
3.74
 
 
8
 
12.32
 
11.17
 
 
9
 
11.42
 
12.81
 
 
10
 
13.61
 
8.95
 
 
11
 
9.41
 
3.36
 
 
12
 
11.08
 
12.47
 
 
13
 
16.39
 
9.70
 
 
14
 
9.83
 
6.26
 
 
15
 
17.65
 
11.26
 
 
16
 
7.57
 
5.92
 
 
17
 
12.73
 
14.83
 
 
18
 
8.82
 
11.89
 
 
19
 
20.12
 
18.35
 
 
20
 
6.67
 
6.09
 
 
21
 
11.36
 
10.25
 
 
22
 
6.33
 
3.86
 
 
23
 
15.09
 
9.11
 
 
24
 
3.27
 
0.63
 
 
25
 
8.72
 
9.20
 
 
26
 
7.72
 
3.86
 
 
27
 
13.31
 
19.70
 
 
28
 
12.43
 
8.57
 
 
29
 
6.96
 
7.60
 
 
30
 
6.16
 
5.88
 
 
Household
Paper
Plastic
Click the icon to view the data.
In this​ example,
μd
is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis​ test?
 
 
A.
H0​:
μd=0
H1​:
μd≠0
 
B.
H0​:
μd=0
H1​:
μd<0
 
C.
H0​:
μd≠0
H1​:
μd>0
 
D.
H0​:
μd≠0
H1​:
μd=0
Identify the test statistic.
 
t=nothing
​(Round to two decimal places as​ needed.)
Identify the​ P-value.
 
​P-value=nothing
​(Round to three decimal places as​ needed.)
What is the conclusion based on the hypothesis​ test?
 
Since the​ P-value is
 
less
greater
than the significance​ level,
 
reject
fail to reject
the null hypothesis. There
 
is not
is
sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.
 
 
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