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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 12 customers and find that the average dollar amount spent per transaction per customer is $116.194 with a standard deviation of $11.3781. Create a 90% confidence interval for the true average spent for all customers per transaction.1) ( 114.398 , 117.99 )2) ( 112.909 , 119.479 )3) ( -110.295 , 122.093 )4) ( 110.341 , 122.047 )5) ( 110.295 , 122.093 )

Sagot :

Answer:

(110.295, 122.093).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 12 - 1 = 11

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.7959

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.7959\frac{11.3781}{\sqrt{12}} = 5.899[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 116.194 - 5.899 = 110.295

The upper end of the interval is the sample mean added to M. So it is 116.194 + 5.899 = 122.093

So

(110.295, 122.093).

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