In: Statistics and Probability
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).
Household Paper Plastic
1 5.86 3.91
2 9.83 6.26
3 9.55 9.20
4 12.43 8.57
5 6.98 2.65
6 11.42 12.81
7 7.57 5.92
8 13.31 19.70
9 6.83 3.57
10 6.38 8.82
11 16.08 14.36
12 8.72 9.20
13 6.33 3.86
14 7.98 6.09
15 6.96 7.60
16 12.32 11.17
17 14.33 6.43
18 11.08 12.47
19 3.27 0.63
20 6.16 5.88
21 2.80 5.92
22 15.09 9.11
23 13.05 12.31
24 13.61 8.95
25 9.41 3.36
26 8.82 11.89
27 9.45 3.02
28 20.12 18.35
29 6.67 6.09
30 9.19 3.74
In this example,
mu Subscript dμd
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 0H0:
mu Subscript dμdequals=0
Upper H 1H1:
mu Subscript dμdnot equals≠0
B.
Upper H 0H0:
mu Subscript dμdequals=0
Upper H 1H1:
mu Subscript dμdless than<0
C.
Upper H 0H0:
mu Subscript dμdnot equals≠0
Upper H 1H1:
mu Subscript dμdgreater than>0
D.
Upper H 0H0:
mu Subscript dμdnot equals≠0
Upper H 1H1:
mu Subscript dμdequals=0
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
▼
less
greater
than the significance level,
▼
fail to reject
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.
(The above calculations are obtained using R-software. The code and output are attached below).