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 0.10 to test for a difference between the weights of discarded paper? (in pounds) and weights of discarded plastic? (in pounds). LOADING... Click the icon to view the data. In this? example, mu Subscript d 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. Upper H 0?: mu Subscript dequals0 Upper H 1?: mu Subscript dless than0 B. Upper H 0?: mu Subscript dnot equals0 Upper H 1?: mu Subscript dequals0 C. Upper H 0?: mu Subscript dequals0 Upper H 1?: mu Subscript dnot equals0 D. Upper H 0?: mu Subscript dnot equals0 Upper H 1?: mu Subscript dgreater than0 Identify the test statistic. tequals nothing ?(Round to two decimal places as? needed.) Identify the? P-value. ?P-valueequals nothing ?(Round to three decimal places as? needed.) What is the conclusion based on the hypothesis? test? Since the? P-value is ? greater less than the significance? level, ? fail to reject reject the null hypothesis. There ? is is not sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.

Household

Paper

Plastic

1

12.3212.32

11.1711.17

2

9.459.45

3.023.02

3

13.3113.31

19.7019.70

4

6.676.67

6.096.09

5

9.559.55

9.209.20

6

6.166.16

5.885.88

7

11.0811.08

12.4712.47

8

15.0915.09

9.119.11

9

12.4312.43

8.578.57

10

9.199.19

3.743.74

11

14.3314.33

6.436.43

12

20.1220.12

18.3518.35

13

6.986.98

2.652.65

14

11.4211.42

12.8112.81

15

12.7312.73

14.8314.83

16

9.419.41

3.363.36

17

5.865.86

3.913.91

18

11.3611.36

10.2510.25

19

2.802.80

5.925.92

20

17.6517.65

11.2611.26

21

13.6113.61

8.958.95

22

8.828.82

11.8911.89

23

13.0513.05

12.3112.31

24

8.728.72

9.209.20

25

6.446.44

8.408.40

26

16.0816.08

14.3614.36

27

9.839.83

6.266.26

28

16.3916.39

9.709.70

29

6.836.83

3.573.57

30

6.336.33

3.863.86

Household

Paper

Plastic

Household Paper Plastic

1 12.32 11.17

2 9.45 3.02

3 13.31 19.70

4 6.67 6.09

5 9.55 9.20

6 6.16 5.88

7 11.08 12.47

8 15.09 9.11

9 12.43 8.57

10 9.19 3.74

11 14.33 6.43

12 20.12 18.35

13 6.98 2.65

14 11.42 12.81

15 12.73 14.83

16 9.41 3.36

17 5.86 3.91

18 11.36 10.25

19 2.80 5.92

20 17.65 11.26

21 13.61 8.95

22 8.82 11.89

23 13.05 12.31

24 8.72 9.20

25 6.44 8.40

26 16.08 14.36

27 9.83 6.26

28 16.39 9.70

29 6.83 3.57

30 6.33 3.86

Household Paper Plastic

Answer 1

let   is the true mean difference between the weight of discarded paper and the weight of discarded plastic for a household.

Using a significance level of 0.10, we want to test for a difference between the weights of discarded paper and weights of discarded plastic

that is we want to test if the difference is equal to zero

The hypotheses that we want to test are

ans: C

Since the samples are paired, we will use dependent sample analysis. The sample size n=30 is less than or equal to 30. Hence this is a small sample analysis. Since the  differences have a distribution that is approximately normal, we will use t-distribution for the analysis.

We calculate the difference d and the sample mean and standard deviation

	paper	plastic	difference (d)
1	12.32	11.17	1.15
2	9.45	3.02	6.43
3	13.31	19.7	-6.39
4	6.67	6.09	0.58
5	9.55	9.2	0.35
6	6.16	5.88	0.28
7	11.08	12.47	-1.39
8	15.09	9.11	5.98
9	12.43	8.57	3.86
10	9.19	3.74	5.45
11	14.33	6.43	7.9
12	20.12	18.35	1.77
13	6.98	2.65	4.33
14	11.42	12.81	-1.39
15	12.73	14.83	-2.1
16	9.41	3.36	6.05
17	5.86	3.91	1.95
18	11.36	10.25	1.11
19	2.8	5.92	-3.12
20	17.65	11.26	6.39
21	13.61	8.95	4.66
22	8.82	11.89	-3.07
23	13.05	12.31	0.74
24	8.72	9.2	-0.48
25	6.44	8.4	-1.96
26	16.08	14.36	1.72
27	9.83	6.26	3.57
28	16.39	9.7	6.69
29	6.83	3.57	3.26
30	6.33	3.86	2.47

Sample mean is

the sample standard deviation of d is

We do not know the population standard deviation of the difference d. We will use sample to estimate the population SD

the standard error of mean difference is

The hypothesized valeu of the mean difference is

The test statistics is

ans: The test statistics t=2.97

The degrees of freedom for t statistics is n-1 = 30-1 =29

This is a 2 tail test. The p-value is the sum of probabilities P(T<-2.97)+P(T>2.97)

Using excel fucntion =T.DIST.2T(2.97,29) we get 0.006

ans: p-value is 0.006

The p-value is less than the the level of significance 0.1

ans: Sicne the p-value is less than the significance level 0.1 we reject the null hypothesis

We conclude that there is sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.