1. A consumer power analyst is hired by a company to investigate whether the pop
ID: 3321541 • Letter: 1
Question
1. A consumer power analyst is hired by a company to investigate whether the popularity of a certain new product depends on the location of the consumers. The monthly averages of the number of the product sold in 20 stores in State A, and 24 stores in State B, are used for the analysis. Conduct a test at 5% significance level to determine whether or not the analyst can claim that there is no difference between the two states State A State B 1791 1823 2070 1737 1630 2079 1944 1998 2034 2066 1908 1888 1755 1958 1854 1823 2034 1670 1868 1755 1797 1999 1823 2039 1602 1499 2120 1868 1944 2093 2075 2008 1746 1801 1962 1814 1782 2061 1647 1890 1746 1828 2111 a. What are the hypotheses (Ho Vs. H1) for the test? b. Determine the critical value for the test c. Compute the statistic for the test d. What is the result of the test? e. Draw a conclusion based on the result of the test f. Determine the 95% confidence interval for the difference of sales between the two statesExplanation / Answer
1.
Given that,
mean(x)=1907.65
standard deviation , s.d1=181.878
number(n1)=20
y(mean)=1862.7083
standard deviation, s.d2 =122.71865
number(n2)=24
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /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-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1907.65-1862.7083/sqrt((33079.60688/20)+(15059.86706/24))
to =0.94
| to | =0.94
critical value
the value of |t | with min (n1-1, n2-1) i.e 19 d.f is 2.09
we got |to| = 0.9409 & | t | = 2.09
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.9409 ) = 0.359
hence value of p0.05 < 0.359,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
c.
test statistic: 0.94
b.
critical value: -2.09 , 2.09
d.
decision: do not reject Ho
p-value: 0.359
e.
we do not have enough evidence to support the claim that there is no difference two states
f.
TRADITIONAL METHOD
given that,
mean(x)=1907.65
standard deviation , s.d1=181.878
number(n1)=20
y(mean)=1862.7083
standard deviation, s.d2 =122.71865
number(n2)=24
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((33079.607/20)+(15059.867/24))
= 47.765
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 19 d.f is 2.093
margin of error = 2.093 * 47.765
= 99.972
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1907.65-1862.7083) ± 99.972 ]
= [-55.03 , 144.913]
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DIRECT METHOD
given that,
mean(x)=1907.65
standard deviation , s.d1=181.878
sample size, n1=20
y(mean)=1862.7083
standard deviation, s.d2 =122.71865
sample size,n2 =24
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 = [( 1907.65-1862.7083) ± t a/2 * sqrt((33079.607/20)+(15059.867/24)]
= [ (44.942) ± t a/2 * 47.765]
= [-55.03 , 144.913]
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interpretations:
1. we are 95% sure that the interval [-55.03 , 144.913] 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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