Question

1. Suppose a researcher, interested in obtaining an estimate of the average level of some enzyme in a certain human population, takes a sample of 36 individuals, determines the level of the enzyme in each, and computes a sample mean of  22. Suppose further it is known that the variable of interest is approximately normally distributed with standard deviation of 7. Give a 94% confidence interval of population mean.

Answer 1

Solution :

Given that,

sample mean = = 22

Population standard deviation =    = 7

Sample size = n =36

At 94% confidence level the z is ,

Z/2 = Z0.03 =1.881 ( Using z table )

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

= 1.881 * (7 /  36 )

= 2.19
At 95% confidence interval of the population mean
is,

- E < < + E

22-2.19 < < 22+2.19

(19.81 ,24.19)