Question

19. A sample of 64 elements from a population with a standard deviation of 48 is selected. The sample mean is 170. You are asked to construct a 95% confidence interval for μ. The lower bound (or lower limit) for that interval is . . . (NOTE: Round your final answer to 2 decimal places - such as 14.12 or 31.00 or 99.55)

10. 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.

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

Answer 1

19)

Solution :

Given that,

Point estimate = sample mean = = 170

Population standard deviation = = 48

Sample size = n = 64

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 * (48 / 64)

= 11.76

At 95% confidence interval estimate of the population mean is,

- E < < + E

170 - 11.76 < < 170 + 11.76

158.24 < < 181.76

Lower bound = 158.24

Upper bound = 181.76

10

Given that,

Point estimate = sample mean = = 141000

Population standard deviation = = 12000

Sample size = 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

Margin of error = 3920

At 95% confidence interval estimate of the population mean is,

- E < < + E

141000 -3920 < < 14100 + 3920

137080 < < 144920