Question

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B, are selected on an experimental basis. Location A was observed for 13 days and location B was observed for 18 days. The number of the particular items sold per day was recorded for each location. On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 55 of these items with a sample standard deviation of 2. Select a 90% confidence interval for the difference in the true means of items sold at location A and B

Answer 1

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

• $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$  

In order to use t*, we need to check conditions for using a t-distribution first.

• Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
• Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
• Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

• $\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

• df for A: 13 - 1 = 12
• df for B: 18 - 1 = 17
• df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

• t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

• $\overline {x}_A=39$
• $s_A=8$
• $n_A=13$
• $\overline {x}_B = 55$
• $s_B=2$
• $n_B=18$
• $t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$.

• $39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

• $-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
• $-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

• $(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.