Question

A random sample of 40 bags of flour is weighed. It is found that the sample mean is 453.08 gram. All weights of the bags of flour have a population standard deviation of 5.42 gm. What is the lower confidence limit (LCL) of an 80% confidence interval for the population mean weight?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 453.08

Population standard deviation =    = 5.42

Sample size n =40

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.10

Z/2 = Z0.10= 1.28 ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.28* ( 5.42/ 40  )

= 1.097
At 80% confidence interval
is,

- E < < + E

453.08 - 1.097 <   < 453.08 + 1.097

451.983 <   < 454.177