Question

Suppose that a sample of 200 accounts receivable entries at a large mail-order firm had a mean price of $846.20 and a standard deviation of $1,840.80. Give a 95% confidence interval for the population mean. Be sure to state any assumptions that you use______________

Answer 1

Answer:

[ $591.08, $1101.32 ]

Explanation:

Given:

Sample space = 200

Mean price = $846.20

Standard deviation, σ = $1,840.80

Confidence level = 95%

Now,

Confidence interval is given as:

⇒ Mean ± $z\frac{\sigma}{\sqrt{n}}$

here, z value for 95% is 1.96 from the standard z table

Thus,

Confidence interval

⇒ $846.20 ± $1.96\times\frac{\ 1,840.80}{\sqrt{200}}$

or

⇒ $846.20 ± $1.96\times\frac{\ 1,840.80}{\sqrt{200}}$

or

⇒ $846.20 ± 255.12

or

⇒ [ $846.20 - 255.12, $846.20 + 255.12 ]

or

⇒  [ $846.20 - 255.12, $846.20 + 255.12 ]

or

⇒ [ $591.08, $1101.32 ]