Question

For each exercise, answer the following along with any additional questions. Assume group variances are equal (unless the problem is ran via statistical software). |  Provide the null and alternative hypotheses in formal and plain language for appropriate two-tailed test (viz., dependent or independent) at the 0.05 significance level  Do the math and reject/accept null at a=.05. State your critical t value.  Explain the results in plain language.  Calculate the 95% confidence interval for the difference in means and state both formally and in plain language if appropriate.

Home Harvest, a local homeless support charity in the northeastern U.S. traditionally mails donation requests to previous donors in the last five years around Thanksgiving. A board member suggests e-mail might be more efficient and cheaper but there is concern an email would be seen as less personal and accrue lower donations. Home Harvest takes a random sample of sixteen regular personal donors (viz., have donated each of the last three years). Six receive an e-mail and eleven a letter for the next Thanksgiving. Donations in dollars are as follows:

E-mail: 40, 0, 60, 40, 70, 0

Letter: 110, 550, 120, 650, 0, 100, 120, 20, 110, 110, 100

Answer 1

a.
Given that,
mean(x)=35
standard deviation , s.d1=29.4958
number(n1)=6
y(mean)=180.9091
standard deviation, s.d2 =212.2477
number(n2)=11
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.131
since our test is two-tailed
reject Ho, if to < -2.131 OR if to > 2.131
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*870.002 + 10*45049.086) / (17- 2 )
s^2 = 30322.725
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=35-180.9091/sqrt((30322.725( 1 /6+ 1/11 ))
to=-145.909/88.376
to=-1.651
| to | =1.651
critical value
the value of |t α| with (n1+n2-2) i.e 15 d.f is 2.131
we got |to| = 1.651 & | t α | = 2.131
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 != -1.651 ) = 0.1182
hence value of p0.05 < 0.1182,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.651
critical value: -2.131 , 2.131
decision: do not reject Ho
p-value: 0.1182
we do not have enough evidence to support the claim that difference in means and state both formally and in plain language
b.
TRADITIONAL METHOD
given that,
mean(x)=35
standard deviation , s.d1=29.4958
number(n1)=6
y(mean)=180.9091
standard deviation, s.d2 =212.2477
number(n2)=11
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*870.002 + 10*45049.086) / (17- 2 )
s^2 = 30322.725
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 30322.725 * (1/6+1/11) )
=88.376
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 15 d.f is 2.131
margin of error = 2.131 * 88.376
= 188.33
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (35-180.9091) ± 188.33 ]
= [-334.239 , 42.421]
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DIRECT METHOD
given that,
mean(x)=35
standard deviation , s.d1=29.4958
sample size, n1=6
y(mean)=180.9091
standard deviation, s.d2 =212.2477
sample size,n2 =11
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 35-180.9091) ± t a/2 * sqrt( 30322.725 * (1/6+1/11) ]
= [ (-145.909) ± 188.33 ]
= [-334.239 , 42.421]
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interpretations:
1. we are 95% sure that the interval [-334.239 , 42.421]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