An analyst is assigned a blind quality control sample of blood containing a drug metabolite. He tests ten separate portions and obtains the following data for the concentration in ug/mL:
25.2
24.8
26.6
27.0
24.0
26.0
25.5
26.3
24.9
23.9
The student t value for 10 samples, nine degrees of freedom is 2.262. Using the sample standard deviation for this data set, what is the ± value of the 95% confidence interval? Report to two decimal places

Question

Quizplus · Accepted Answer

To calculate the 95% confidence interval, we first need to find the standard error of the mean (SEM). 
SEM = sample standard deviation / sqrt(n)
where n = sample size (in this case, 10)

Using the given data, we find the sample standard deviation to be 1.12. 
SEM = 1.12 / sqrt(10) = 0.35

Next, we need to find the margin of error (ME). 
ME = t*SEM
where t = student t value with n-1 degrees of freedom and desired level of confidence (in this case, 95%), which is given as 2.262.

ME = 2.262 * 0.35 = 0.79

Finally, we can calculate the confidence interval (CI) by adding and subtracting the margin of error from the sample mean.

CI = sample mean ± ME
where the sample mean is the average concentration of the 10 portions:
(25.2 + 24.8 + 26.6 + 27.0 + 24.0 + 26.0 + 25.5 + 26.3 + 24.9 + 23.9) / 10 = 25.2

CI = 25.2 ± 0.79
CI = (24.41, 25.99)

Therefore, the answer is C (0.76) as it is the closest option to the calculated margin of error of 0.79.

An Analyst Is Assigned a Blind Quality Control Sample of Blood