1/8÷2= as mixed # if possible
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Answer:
We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.
Step-by-step explanation:
We are given that Location A was observed for 18 days and location B was observed for 13 days.
On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.
<em>Let </em><em> = true mean number of sales at location A.</em>
<em /> = <em>true mean number of sales at location B</em>
So, Null Hypothesis, :
or
{means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}
Alternate Hypothesis, :
or
{means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}
The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;
T.S. = ~
where, = sample average of items sold at location A = 39
= sample average of items sold at location B = 49
= sample standard deviation of items sold at location A = 8
= sample standard deviation of items sold at location B = 4
= sample of days location A was observed = 18
= sample of days location B was observed = 13
Also, =
= 6.64
So, <u><em>test statistics</em></u> = ~
= -4.14
The value of t test statistics is -4.14.
Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.
<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.