Question 1193373: The daily sales of a product in different shops follow a normal distribution. The targeted mean of daily sales of the product and their variance in the population are k10 , 000 and k2 500 respectively. The sales manager of the company which produces the product feels that the sales of the product has improved in the recent past. A random sample of 49 shops is taken from the normal population for which the mean daily sales is found to be k 9 600. Check whether the sales of the product in different shops has really improved at a significant level of 0.05
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I am assuming the mean daily sales is 9600 with sd of 50 (sqrt(2500)).
Ho: the mean of the population is 10,000
Ha: the mean is not 10,000
alpha=0.05 p{reject Ho|Ho true}
test stat is z=(x bar-mean)/sigma/sqrt(n)
critical value is |z|>1.96
z=(9600-10000)/50/sqrt(49)
=-400*7/50
=-56 with p-value of 0.
Reject Ho: the mean is not 10000; it is smaller.
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I am unclear about the targeted mean and variance. If one used 2500 as the sd, then z would have been -1.12, at least more reasonable an answer than -56. The units of the variance are k^2, so if k denotes units, then what is being given is the standard deviation, not the variance.
The question asks whether the product has really improved, but there is no prior value. The hypothesis test above basically says if the population were 10000 and the variance 2500, what would be the likelihood that a sample of 49 would have a mean of 9600? That probability is 0.
If this does not fit with what was desired, please resubmit.
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