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.
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.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
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Topic Video
Question
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,
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?
μd
is the H0:
μd=0
H1:
μd≠0
H0:
μd=0
H1:
μd<0
H0:
μd≠0
H1:
μd>0
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
than the significance level,
the null hypothesis. There
sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.
▼
less
greater
▼
reject
fail to reject
▼
is not
is
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