The output is one-fifth of the input
1 answer:
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Answer:
We conclude that there is no difference in potential mean sales per market in Region 1 and 2.
Step-by-step explanation:
We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.
A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.
Let = mean sales per market in Region 1.
= mean sales per market in Region 2.
So, Null Hypothesis, :
= 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}
Alternate Hypothesis, : >
0 {means that there is a difference in potential mean sales per market in Region 1 and 2}
The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;
T.S. = ~
where, = sample mean sales in Region 1 = 84
= sample mean sales in Region 2 = 78.3
= sample standard deviation of sales in Region 1 = 6.6
= sample standard deviation of sales in Region 2 = 8.5
= sample of supermarkets from Region 1 = 12
= sample of supermarkets from Region 2 = 17
Also, =
= 7.782
So, <u><em>the test statistics</em></u> = ~
= 1.943
The value of t-test statistics is 1.943.
Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.
Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.
Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.