Grande-Terre m = 10 a:1= 64.09 S16.5g Malabar n2 = 10 x2=82.7g 82= 3.69 (a) Comp

Question

Grande-Terre m = 10 a:1= 64.09 S16.5g Malabar n2 = 10 x2=82.7g 82= 3.69 (a) Compute a 95% confidence interval for the difference in mean weights of eggs at the tine of lay on Malabar Island and Grand-Terre Island. (b) Do the results of this survey support he proposed contention? i. Using standard notation, define the population parameter(s) being tested ii. Specify the null and alternative hypotheses. iii. Compute the observed value of the test statistic. iv. Specify the distribution to be used for computing the p-value and compute the p- value within table accuracy v. State your conclusion and report the estimated value of the parameter being tested and the (estimated) standard error vi. What assumptions have you made to carry out this test? Use these results as a pilot study to determine the common sample size needed to estimate the difference in mean weights of eggs at the time of lay on Malabar Island and Grand- Terre Island to within 1 g with 95% confidence. (c)

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
mean(x)=64
standard deviation , s.d1=6.5
number(n1)=10
y(mean)=82.7
standard deviation, s.d2 =3.6
number(n2)=10
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((42.25/10)+(12.96/10))
= 2.35
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
margin of error = 2.262 * 2.35
= 5.315
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (64-82.7) ± 5.315 ]
= [-24.015 , -13.385]
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DIRECT METHOD
given that,
mean(x)=64
standard deviation , s.d1=6.5
sample size, n1=10
y(mean)=82.7
standard deviation, s.d2 =3.6
sample size,n2 =10
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 = [( 64-82.7) ± t a/2 * sqrt((42.25/10)+(12.96/10)]
= [ (-18.7) ± t a/2 * 2.35]
= [-24.015 , -13.385]
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interpretations:
1. we are 95% sure that the interval [-24.015 , -13.385] 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

PART B.
Given that,
mean(x)=64
standard deviation , s.d1=6.5
number(n1)=10
y(mean)=82.7
standard deviation, s.d2 =3.6
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /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-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =64-82.7/sqrt((42.25/10)+(12.96/10))
to =-7.9585
| to | =7.9585
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 7.95853 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -7.9585 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------

i.

population paramater, mean weights of eggs at the time of lay

ii.
null, Ho: u1 = u2
alternate, H1: u1 != u2
iii.
test statistic: -7.9585
iv.
p-value: 0
v.
decision: reject Ho
vi.
there is diffrence in mean weights of eggs at the time of lay on malabar iland

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