Question

To determine the weight of plastic discarded by households, a sample size of 62 weights are measured and are found to have a mean of 1.911 lb and a standard deviation of 1.065 lb. Construct a 99% confidence interval estimate of the mean weight of plastic discarded by all households.

Answer 1

Solution :

Given that,

= 1.911

s = 1.065

n = 62

Degrees of freedom = df = n - 1 = 62 - 1 = 61

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,61 = 2.659

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

= 2.659 * (1.065 / 62)

= 0.360

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

- E < < + E

1.911 - 0.360 < < 1.911 + 0.360

1.551 < < 2.271

(1.551, 2.271)