In: Statistics and Probability
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.
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 =
t0.005,61 = 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)