Question

Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected on an experimental basis. Location A was observed for 18 days and location B was observed for 13 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 49 of these items with a sample standard deviation of 4. Select a 99% confidence interval for the difference in the true means of items sold at location A and B. a) O [-1242,-7582]

b) O132.76, 45.24]

c)。8 1.76, 94.24]

d) 0-1 6.03,-3.97]

e)。[42.76, 55.24]

F. None of the above

Answer 1

Answer:

d) [-16.03,-3.97]

$-16.03 \leq \mu_A -\mu_B \leq -3.97$.

Step-by-step explanation:

Notation and previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$n_A=18$ represent the sample of A

$n_B =13$ represent the sample of B

$\bar x_A =39$ represent the mean sample  for A

$\bar x_B =49$ represent the mean sample for B  

$s_A =8$ represent the sample deviation for A

$s_B =4$ represent the sample deviation for B

$\alpha=0.01$ represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$ (1)  

The point of estimate for $\mu_A -\mu_B$ is just given by:  

$\bar X_A -\bar X_B =39-49=-10$  

The appropiate degrees of freedom are $df=n_1+ n_2 -2=18+13-2=29$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that $t_{\alpha/2}=2.756$  

The standard error is given by the following formula:  

$SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$  

And replacing we have:  

$SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188$  

Confidence interval  

Now we have everything in order to replace into formula (1):  

$-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03$  

$-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97$  

So on this case the 99% confidence interval for the differences of means would be given by $-16.03 \leq \mu_A -\mu_B \leq -3.97$.

d) [-16.03,-3.97]