Question

a.

Find and interpret a 95% confidence interval based on the following:

65 out of 83 sampled students believe that their lives will be forever altered by the coronavirus.

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b.

In an effort to compare vegetation growth in an area a cluster sample was done in 2006 at 8 locations each 1 square foot and mapped with GPS coordiantes.

The next year the researchers came back to the same area to see if vegetation had increased. They used the coordinates and mapped the same square foot units. The data is in the table below. Test to see if vegetation has increased in 2007 from the year previous.  

	2006	2007
Location A	124	132
Location B	35	56
Location C	21	12
Location D	245	270
Location E	151	180
Location F	78	81
Location G	95	11
Location H	82	96

Complete all 8 steps of a hypothesis test. You may skip making a histogram.  

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=65
sample size(n)=83
success rate ( p )= x/n = 0.7831
I.
sample proportion = 0.7831
standard error = Sqrt ( (0.7831*0.2169) /83) )
= 0.0452
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.0452
= 0.0887
III.
CI = [ p ± margin of error ]
confidence interval = [0.7831 ± 0.0887]
= [ 0.6945 , 0.8718]
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DIRECT METHOD
given that,
possible chances (x)=65
sample size(n)=83
success rate ( p )= x/n = 0.7831
CI = confidence interval
confidence interval = [ 0.7831 ± 1.96 * Sqrt ( (0.7831*0.2169) /83) ) ]
= [0.7831 - 1.96 * Sqrt ( (0.7831*0.2169) /83) , 0.7831 + 1.96 * Sqrt ( (0.7831*0.2169) /83) ]
= [0.6945 , 0.8718]
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interpretations:
1. We are 95% sure that the interval [ 0.6945 , 0.8718] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
b.
Given that,
mean(x)=104.75
standard deviation , s.d1=87.9298
number(n1)=8
y(mean)=103.875
standard deviation, s.d2 =71.0944
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =104.75-103.875/sqrt((7731.64973/8)+(5054.41371/8))
to =0.0219
| to | =0.0219
critical value
the value of |t α| with min (n1-1, n2-1) i.e 7 d.f is 1.895
we got |to| = 0.02189 & | t α | = 1.895
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.0219 ) = 0.49157
hence value of p0.05 < 0.49157,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 0.0219
critical value: 1.895
decision: do not reject Ho
p-value: 0.49157
we do not have enough evidence to support the claim that if vegetation has increased in 2007 from the year previous