Question

A local electronics firm wants to determine their average daily sales (in dollars.) A sample of the sales for 36 days revealed an average sales of $141,000. Assume that the standard deviation of the population is known to be $12,000. Then, the 95% margin of error will be . . .

Provide a 95% confidence interval estimate for the true average daily sales.

	$137,080 to $144,920
	$139,500 to $143,000
	$132,800 to 139,200
	$136,200 to $142,200
	$140,600 to $143,200
	$135,600 to $143,400
	$140,160 to $141,840

Answer 1

Solution :

Given that,

= 141000

= 12000

n = 36

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (12000 / 36)

= 3920

At 99% confidence interval estimate of the population mean is,

- E < < + E

141000-3920< < 141000+3920

137080< < 144920

($137,080 to $144,920)

correct option (a)$137,080 to $144,920