A company is trying out a new marketing plan. Prior to the new campaign, store s
ID: 3160321 • Letter: A
Question
A company is trying out a new marketing plan. Prior to the new campaign, store sales were $5000 per week. The new method is trialled in 81 randomly selected stores and results in mean sales of $5245 and standard deviation of $1100.
(a) Is there evidence at the 0.05 level of significance to conclude that the new marketing plan works? Show your working and clearly indicate the assumptions/conditions required to perform the test.
(b) How large a sample must be selected if the company wants to be 95% confident that the maximum error of estimation is $154.
Explanation / Answer
a)
Here, we assume that the underlying distribution is approximately normally distributed and has no outliers.
Formulating the null and alternative hypotheses,
Ho: u <= 5000
Ha: u > 5000
As we can see, this is a right tailed test.
Thus, getting the critical t,
df = n - 1 = 80
tcrit = + 1.664124579
Getting the test statistic, as
X = sample mean = 5245
uo = hypothesized mean = 5000
n = sample size = 81
s = standard deviation = 1100
Thus, t = (X - uo) * sqrt(n) / s = 2.004545455
Also, the p value is
p = 0.02419743
As t > 1.664, and P < 0.05, we REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that the new marketing plan works. [CONCLUSION]
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b)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
s = sample standard deviation = 1100
E = margin of error = 154
Thus,
n = 195.992797
Rounding up,
n = 196 [ANSWER]
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