Question

The sanitation department of a large city wants to investigate ways to reduce the amount of recyclable materials that are placed in the landfill. They randomly sample 30 households and find that the sample mean amount of waste per household is 9.5 pounds/week and the standard deviation is 11.5 pounds/week. The distribution of the data is slightly right skewed but contains no extreme outliers. Construct a 99% confidence interval for the mean amount of waste per household.

	a.	(5.385, 13.615)
	b.	The conditions are not met because the data distribution is not perfectly normal.
	c.	(3.726, 15.274)
	d.	(4.091, 14.909)
	e.	(3.713, 15.287)

Answer 1

Solution :

Given that,

= 9.5

s = 11.5

n =30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,29 =2.756

Margin of error = E = t/2,df * (s /n)

= 2.756 * (11.5 / 30)

= 5.787

Margin of error =5.787

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

- E < < + E

9.5 - 5.787 < < 9.5 + 5.787

3.713 < < 15.287

(3.713, 15.287)

Option e ) is correct.