Question

You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 26 customers and find that the average dollar amount spent per transaction per customer is $77.506 with a standard deviation of $11.0714. When creating a 99% confidence interval for the true average dollar amount spend per customer, what is the margin of error

Answer 1

Answer:

Step-by-step explanation:

Hello!

The Confidence intervals for the population mean to follow the structure "point estimation"±" margin of error"

You as the store owner took a sample to determine the average amount of money a typical customer spends on your shop.

n= 26 customers.

X[bar]= $77.506

S= $11.0714

Assuming the variable X: the amount of money spent by a typical customer has a normal distribution and that there is no information about the population standard deviation, the best statistic to use for this estimation is a Students-t:

[X[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$]

$t_{n-1; 1-\alpha /2}= t_{25; 0.995}= 2.787$

The margin of error of the interval is:

d= $t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

$d= 2.787 * \frac{11.0714}{\sqrt{26} }$

d= 6.051

I hope it helps!