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

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] ± * ]

The margin of error of the interval is:

d= *

d= 6.051

I hope it helps!

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.