In: Statistics and Probability
Zombies primary food source is brains and spinal cords--a diet high in protein and low in carbohydrates. Serum cholesterol levels are measured in 25 randomly selected adult zombies. The mean of the sample is found to be 151.4.
Sample Mean = 151.4 mg/dl
Sample Standard Deviation = 38 mg/dl.
Sample Size = 25
Calculate the lower value for the 95% confidence interval for the mean cholesterol level for all zombies when the population standard deviation is not known and it estimated by the SD of the sample
Solution :
Given that,
=154.4
s = 38
n = 25
Degrees of freedom = df = n - 1 = 25 - 1 = 24
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,24 = 2.064
Margin of error = E = t/2,df * (s /n)
= 2.064 * (38 / 25)
= 15.7
The 95% confidence interval estimate of the population mean is,
- E < < + E
151.4 - 15.7 < < 151.4 + 15.7
135.7 < < 167.1
(135.7 , 167.1)