Damage to grapes from bird predation is a serious problem for grape growers. An

Question

Damage to grapes from bird predation is a serious problem for grape growers. An article reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time (sec) spent on a single visit to the location. (Note: Assume the populations are normally distributed.) Species Blackbirds Blackbirds Silvereyes Silvereyes Location Exptl Natural Exptl Natural 13.7 9.7 49.1 38.3 SE mean 2.04 1.78 4.75 5.09 35 29 (a) Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location. (Use = 0.05. Round your answer to two decimal places.) sec (b) Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses. (Use = 0.05.) State the relevant hypotheses. (Use for blackbirds at the experimental location and 2 for blackbirds at the natural location.) Compute the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) P-value - State the conclusion in the problem context

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample mean, x =13.7
standard deviation, s =2.04
sample size, n =40
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.04/ sqrt ( 40) )
= 0.32
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 39 d.f is 2.023
margin of error = 2.023 * 0.32
= 0.65
III.
CI = x ± margin of error
confidence interval = [ 13.7 ± 0.65 ]
= [ 13.05 , 14.35 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =13.7
standard deviation, s =2.04
sample size, n =40
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 39 d.f is 2.023
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 13.7 ± t a/2 ( 2.04/ Sqrt ( 40) ]
= [ 13.7-(2.023 * 0.32) , 13.7+(2.023 * 0.32) ]
= [ 13.05 , 14.35 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 13.05 , 14.35 ] contains the true population mean
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 mean
upper confidence interval = 14.35 sec
b.
Given that,
mean(x)=13.7
standard deviation , s.d1=2.04
number(n1)=40
y(mean)=9.7
standard deviation, s.d2 =1.78
number(n2)=35
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.69
since our test is right-tailed
reject Ho, if to > 1.69
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =13.7-9.7/sqrt((4.1616/40)+(3.1684/35))
to =9.068
| to | =9.068
critical value
the value of |t | with min (n1-1, n2-1) i.e 34 d.f is 1.69
we got |to| = 9.06832 & | t | = 1.69
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 9.0683 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 9.068 = 9.07
critical value: 1.69
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference between blackbirds experimental location and blackbirds natural location
c.
TRADITIONAL METHOD
given that,
mean(x)=38.3
standard deviation , s.d1=5.09
number(n1)=29
y(mean)=9.7
standard deviation, s.d2 =1.78
number(n2)=35
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((25.91/29)+(3.17/35))
= 0.99
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 28 d.f is 2.05
margin of error = 2.048 * 0.99
= 2.03
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (38.3-9.7) ± 2.03 ]
= [26.57 , 30.63]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=38.3
standard deviation , s.d1=5.09
sample size, n1=29
y(mean)=9.7
standard deviation, s.d2 =1.78
sample size,n2 =35
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 38.3-9.7) ± t a/2 * sqrt((25.91/29)+(3.17/35)]
= [ (28.6) ± t a/2 * 0.99]
= [26.57 , 30.63]sec
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [26.57 , 30.63] 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

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