5:25 PM webwork.gonzaga.edu AT&T; 80% D + Prev Up Next (1 pt) A market research

Question

5:25 PM webwork.gonzaga.edu AT&T; 80% D + Prev Up Next (1 pt) A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores. Marketing managers are prone to look at the estimate and ignore sampling error. An SRS of 19 stores this year shows mean sales of 58 units of a small appliance, with a standard deviation of 5.8 units. During the same point in time last year, an SRS of 12 stores had mean sales of 52.374 units, with standard deviation 17.3 units. An increase from 52.374 to 58 is a rise of about 10% I. Construct a 95% confidence interval estimate of the difference 1-112, where 1 is the mean of this year's sales and 112 is the mean of last year's sales. (b) The margin of error is 2. At a 0.05 significance level, is there sufficient evidence to show that sales this year are different from last year? A. Yes O B. No

Explanation / Answer

1.
TRADITIONAL METHOD
given that,
mean(x)=58
standard deviation , 1 =5.8
population size(n1)=19
y(mean)=52.374
standard deviation, 2 =17.3
population size(n2)=12
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((33.64/19)+(299.29/12))
= 5.168
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 5.168
= 10.13
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (58-52.374) ± 10.13 ]
= [-4.504 , 15.756]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=58
standard deviation , 1 =5.8
number(n1)=19
y(mean)=52.374
standard deviation, 2 =17.3
number(n2)=12
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 58-52.374) ±Z a/2 * Sqrt( 33.64/19+299.29/12)]
= [ (5.626) ± Z a/2 * Sqrt( 26.711) ]
= [ (5.626) ± 1.96 * Sqrt( 26.711) ]
= [-4.504 , 15.756]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-4.504 , 15.756] contains the difference between
true population mean U1 - U2
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 difference between
true population mean U1 - U2
3. Since this Cl does n't contain a zero we can conclude at 0.05 true mean
difference is not zero
a.
confidence interval = [-4.504 , 15.756]
b.
margin of error = 1.96 * 5.168
= 10.13
2.
Given that,
mean(x)=58
standard deviation , 1 =5.8
number(n1)=19
y(mean)=52.374
standard deviation, 2 =17.3
number(n2)=12
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=58-52.374/sqrt((33.64/19)+(299.29/12))
zo =1.089
| zo | =1.089
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =1.089 & | z | =1.96
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.089 ) = 0.27635
hence value of p0.05 < 0.27635,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 1.089
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.27635

No,
there is sufficient evidence to show that sales this year different from last year

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