Question

1) Consider a survey of garbage production. In a sample of N households, the average household generates 44 gallons of non-recyclable, non-compostable garbage per week, with a standard deviation of 23. The city's landfill is large enough to accommodate 50 gallons of garbage per week per household, and we are interesting in whether we can rule out, using a 95% confidence level, the hypothesis that the landfill is inadequate to the need. a) Is a one-sided or two-sided hypothesis test appropriate here? b) Suppose that the survey sample size is N=10. What is the appropriate critical value (i.e., cutoff in the relevant distribution) to use here? What is the p-value for the test of the hypothesis? c) Suppose that the survey sample size is N=20. What is the appropriate critical value (i.e., cutoff in the relevant distribution) to use here? What is the p-value for the test of the hypothesis? d) Suppose that the survey sample size is N=50. What is the appropriate critical value (i.e., cutoff in the relevant distribution) to use here? What is the p-value for the test of the hypothesis? e) Suppose that the survey sample size is N=100. What is the appropriate critical value (i.e., cutoff in the relevant distribution) to use here? What is the p-value for the test of the hypothesis? f) Suppose that the survey sample size is N=1000. What is the appropriate critical value (i.e., cutoff in the relevant distribution) to use here? What is the p-value for the test of the hypothesis?

Answer 1

1.
a.
two tailed test
b.
Given that,
population mean(u)=50
sample mean, x =44
standard deviation, s =23
number (n)=10
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =44-50/(23/sqrt(10))
to =-0.8249
| to | =0.8249
critical value
the value of |t α| with n-1 = 9 d.f is 2.262
we got |to| =0.8249 & | t α | =2.262
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.8249 ) = 0.4307
hence value of p0.05 < 0.4307,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: -0.8249
b.
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.4307
we do not have enough evidence to support the claim that mean is large enough 50 gallons of garbage.
per week per house hold.

c.
sample size
N = 20
Given that,
population mean(u)=50
sample mean, x =44
standard deviation, s =23
number (n)=20
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.09
since our test is two-tailed
reject Ho, if to < -2.09 OR if to > 2.09
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =44-50/(23/sqrt(20))
to =-1.167
| to | =1.167
critical value
the value of |t α| with n-1 = 19 d.f is 2.09
we got |to| =1.167 & | t α | =2.09
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.1666 ) = 0.2578
hence value of p0.05 < 0.2578,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: -1.167
c.
critical value: -2.09 , 2.09
decision: do not reject Ho
p-value: 0.2578

d.
sample size
N=50
Given that,
population mean(u)=50
sample mean, x =44
standard deviation, s =23
number (n)=50
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.01
since our test is two-tailed
reject Ho, if to < -2.01 OR if to > 2.01
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =44-50/(23/sqrt(50))
to =-1.845
| to | =1.845
critical value
the value of |t α| with n-1 = 49 d.f is 2.01
we got |to| =1.845 & | t α | =2.01
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.8446 ) = 0.0711
hence value of p0.05 < 0.0711,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: -1.845
d.
critical value: -2.01 , 2.01
decision: do not reject Ho
p-value: 0.0711

e.
sample size
N =100
Given that,
population mean(u)=50
sample mean, x =44
standard deviation, s =23
number (n)=100
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =1.98
since our test is two-tailed
reject Ho, if to < -1.98 OR if to > 1.98
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =44-50/(23/sqrt(100))
to =-2.609
| to | =2.609
critical value
the value of |t α| with n-1 = 99 d.f is 1.98
we got |to| =2.609 & | t α | =1.98
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.6087 ) = 0.0105
hence value of p0.05 > 0.0105,here we reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: -2.609
e.
critical value: -1.98 , 1.98
decision: reject Ho
p-value: 0.0105

f.
N= 1000
Given that,
population mean(u)=50
sample mean, x =44
standard deviation, s =23
number (n)=1000
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =1.96
since our test is two-tailed
reject Ho, if to < -1.96 OR if to > 1.96
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =44-50/(23/sqrt(1000))
to =-8.249
| to | =8.249
critical value
the value of |t α| with n-1 = 999 d.f is 1.96
we got |to| =8.249 & | t α | =1.96
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -8.2494 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: -8.249
f.
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that mean is 50 gallons per week per house hold.