Question

An article in the New England Journal of Medicine reported that among adults living in the US, the average level of albumin in cerbrospinal fluid is 29.5 mg/dl, with a standard deviation of 9.25 mg/dl. We are going to select a sample of size 20 from this population.

a. How does the variability of our sample mean compare to the variability of albumin levels in the popoulation?

b. What is the probability that our sample mean will be greater than 33 mg/dl?

c. What is the probability that our sample mean will lie between 29 and 31 mg/dl?

d. Suppose we replicate the study, collecting many samples of size 20. What two values will contain the middle 50% of our sample mean?

e. Suppose we collect a random sample from an adolescent population instead and we observe a sample mean of 25 mg/dl. Form a 95% confidence interval for the true mean albumin level in the adolescent population.

f. Based on part c, would you say that it is plausible that there is a relationship between albumin levels and age?

Answer 1

Solution:

Given that population mean, = 29.5, standard deviation, = 9.25 and sample size, n = 20

a. The variability of the sample mean is the standard error, Se = = 9.25/20 = 2.068

b. The respective Z-score with = 33 is

Z =

Z = (33 - 29.5)/2.068

Z = 1.69

Using Z-tables, the probability is

P [Z > 1.69] = 1 - P (Z 1.69)

= 1 - 0.9545

= 0.0455

c. The respective Z-score with = 29 is

Z =

Z = (29 - 29.5)/2.068

Z = -0.24

The respective Z-score with = 31 is

Z =

Z = (31 - 29.5)/2.068

Z = 0.73

Using Z-tables, the probability is

P [-0.24 < Z < 0.73] = P (Z 0.73) - P (Z -0.24)

= 0.7673 - 0.4052

= 0.3621

d. Using Z-tables, the respective z-score with p = 0.50 is -0.674 and 0.674.

We have

Z =

-0.674 = ( - 29.5)/2.068

= 29.5 - 0.674*2.068

= 28.1

Again

Z =

0.674 = ( - 29.5)/2.068

= 29.5 + 0.674*2.068

= 30.9

e. Assuming the population standard deviation is known, we use z-test.

Using z-tables, the critical value at a/2 = 0.025 is 1.96

95% confidence interval is given by:-

25 1.96*9.25/20

25 1.96*2.068

25 4.05

20.95, 29.05

f. Since the probability is not very low, therefore it is plausible that there is a relationship between albumin levels and age.