Question

4. 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 a variance of 45. Give a 96% confidence interval of population mean 5. Suppose we find that, the average engine size in our sample of size n 36 is 192.67 cubic inches, with a standard deviation of 87.55 cubic inches. Use these statistics to compute a 90% confidence interval of population mean, that is. the average engine size for all.

Answer 1

4) 90 % Confidence Interval

Calculation

M = 22
Z= 1.64
s_M = ?(6.0782²/36) = 1.01

? = M ± Z(s_M)
? = 22 ± 1.64*1.01
? = 22 ± 1.67

Result

M = 22, 90% CI [20.33, 23.67].

5)   90 % Confidence Interval

Calculation

M = 192.67
t = 1.64
s_M = ?(87.55²/36) = 14.59

? = M ± Z(s_M)
? = 192.67 ± 1.64*14.59
? = 192.67 ± 24.0012

Result

M = 192.67, 90% CI [168.6688, 216.6712]