The average spending at Neco's salad bar is $9.03 with a standard deviation of $3.26. The distribution follows t-distribution. The management is interested in the middle 90% of the customers (spending wise) as it believes that they represent their true customer base. What will be the difference between the upper and lower spending cut-offs which define the middle 90% of the customers if the sample contains 41 customers

The difference between these cut-offs is of $1.9.

Step-by-step explanation:

In this question, we have to find the 90% confidence interval, using the t-distribution.

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 = 41 - 1 = 40

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 40 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.684

The margin of error is:

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 9.03 - 0.95 = $8.08.

The upper end of the interval is the sample mean added to M. So it is 9.03 + 0.95 = $9.98.

What will be the difference between the upper and lower spending cut-offs which define the middle 90% of the customers if the sample contains 41 customers

$9.98 - $8.08 = $1.9

The difference between these cut-offs is of $1.9.

associative property hopes this ! : )

step-by-step explanation:

i dont know

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