Question

Subject	Vitamin D
1	57.704
2	48.321
3	56.444
4	59.315
5	52.361
6	62.644
7	54.918
8	48.405
9	53.112
10	48.596
11	44.253
12	54.611
13	52.675
14	53.014
15	57.334
16	53.471
17	51.682
18	58.367
19	49.117
20	54.089
21	55.066
22	52.076
23	53.669
24	49.429
25	53.000
26	56.782
27	53.399
28	48.390
29	59.187

What is the estimated 95% confidence interval (CI) of the average blood vitamin D level of US landscapers in ng/mL?

Please note the following: 1) you might calculate a CI that is different from any of the multiple choice options listed below due to rounding differences, therefore select the closest match; 2) ensure you use either the large or small sample CI formula as appropriate; and 3) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. 45.7 to 49.5 ng/mL

b. 46.5 to 51.5 ng/mL

c. 51.9 to 55.0 ng/mL

d. 47.8 to 62.2 ng/mL

Answer 1

First enter Data into EXCEL

We have to find the sample mean.

Excel command is =AVERAGE(Select data)

sample mean=$\bar{x}$ = 53.50

Now we have to find sample standard deviation.

Excel command is =STDEV(Select data)

standard deviation = s = 4.07

n= 29

$\bar{x}$= 53.50

s = 4.07

95 % confidence interval

Formula

$(\bar{x}-E , \bar{x}+E)$

E=

tc = 2.048 ( using t-table )

$E = \frac{2.048*4.07}{\sqrt{29}}$

95 % confidence interval is

(51.9, 55)

option c is correct

c. 51.9 to 55.0 ng/mL