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Twenty-five blood samples were selected by taking every seventh blood sample from racks holding 187 blood samples from the morning draw at a medical center. The white blood count (WBC) was measured using a Coulter Counter Model S. The mean WBC was 8.636 with a standard deviation of 3.9265. (a) Construct a 90 percent confidence interval for the true mean. (b) Why might normality be an issue here? (c) What sample size would be needed for an error of ± 1.5 with 98 percent confidence?

Here it is given that

Sample mean, x-bar = 8.636

Sample standard deviation, s = 3.9265

Sample size, n = 25

t-value = 1.711 [From Student’s t table with n -1 = 25 – 1 = 24 degrees of freedom with ( = 0.10]

Margin of error, E = t*s/√n = 1.711* (3.9265/(25)= 1.3436

Since the sample size is small (size < 30) we will use the t-distribution to calculate the confidence interval.

a) The 90% confidence interval for the true mean is

(x-bar – E, x-bar + E) = (8.636 - 1.3436, 8.636 + 1.3436)

= (7.2924, 9.9796)

(b) Normality might be an issue because the sample size is small and we have no idea how the population is distributed.

(c)

Here it is given that,

[pic] = 3.9265, E = ± 1.5, [pic]= 2.326

Now, the required sample size,

n ≥ [pic]= (2.326*3.9265/1.5)^2 =37.0832

Thus the required sample size = 38

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