Question

Recall in our discussion of the normal distribution the research study that examined the blood vitamin D levels of the entire US population of landscape gardeners. The intent of this large-scale and comprehensive study was to characterize fully this population of landscapers as normally distributed with a corresponding population mean and standard deviation, which were determined from the data collection of the entire population.

Suppose you are now in a different reality in which this study never took place though you are still interested in studying the average vitamin D levels of US landscapers. In other words, the underlying population mean and standard deviation are now unknown to you. Furthermore, you would like to examine if wearing tank tops instead of short sleeve shirts significantly effects vitamin D levels. To accomplish this, you propose to collect data from the landscapers at two different points in time. Specifically, the landscapers are to wear short sleeve shirts while outside working during a period of three weeks. After three weeks, you collect blood specimens and the landscapers are then to wear tank tops for the next three weeks under the same working conditions, after which you collect blood draws a second time. You obtain research funding to randomly sample 35 landscapers, collect blood samples at two different time points as described above, and send these samples to your collaborating lab in order to quantify the amount of vitamin D in the landscapers' blood. After anxiously awaiting your colleagues to complete their lab quantification protocol, they email you the following vitamin D level data as shown in the following table.

Subject	Time Point 1, Shirts Vitamin D (ng/mL)	Time Point 2, Tank Tops Vitamin D (ng/mL)
1	29.010	46.053
2	36.171	48.732
3	30.359	49.307
4	28.864	53.528
5	24.936	47.638
6	30.083	46.050
7	36.513	35.802
8	29.222	47.210
9	32.998	45.908
10	36.204	45.805
11	40.637	45.282
12	33.057	54.345
13	30.536	52.712
14	34.908	49.563
15	32.705	49.026
16	32.155	46.914
17	31.319	47.930
18	43.032	51.729
19	34.367	41.464
20	38.465	56.488
21	33.415	43.805
22	34.214	45.417
23	34.192	39.355
24	35.125	49.289
25	33.143	51.823
26	42.740	45.327
27	44.913	50.023
28	31.256	41.949
29	30.598	46.009
30	27.896	50.630
31	32.304	46.243
32	28.441	43.802
33	33.903	47.878
34	35.446	51.888
35	32.977	49.664

What is the estimated 95% confidence interval (CI) of the average difference in blood vitamin D levels between short sleeve shirt and tank top attire amongst US landscapers in ng/mL?

Please note the following: 1) in practice, you as the analyst decide how to calculate the difference in vitamin D levels between time points for a given study participant, and subsequently interpret the aggregated results appropriately in the context of the data, though for the purposes of this exercise the difference is assigned for you as follows. Define the difference as the second minus the first time points, which is common practice, since the plus or minus sign of the resulting difference reflects any change over sequential time; 2) 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; 3) ensure you use either the large or small sample CI formula as appropriate; and 4) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. 12.80 to 17.06 ng/mL

b. 10.71 to 17.62 ng/mL

c. 11.96 to 15.95 ng/mL

d. 13.28 to 18.50 ng/mL

Answer 1

Step.1 First Copy and paste data in Excel sheet.

Step.2 Find difference as the second minus first time points.

Step.3 Go to 'Data' menu ---> Select 'Data Analysis'. New window will pop-up on screen.

Step.4 Select descriptive statistics option.

Step.5 Select values in difference column as input range. Also mention output range. You will get the following answer.

Column1
Mean	13.95668571
Standard Error	1.016246303
Median	14.759
Mode	#N/A
Standard Deviation	6.01219421
Sample Variance	36.14647922
Kurtosis	-0.044190481
Skewness	-0.497325552
Range	25.375
Minimum	-0.711
Maximum	24.664
Sum	488.484
Count	35
Confidence Level(95.0%)	2.065260959

95% confidence interval :

Lower limit = mean - Confidence Level(95.0%) = 13.96 - 2.07 = 11.89

Upper limit = mean + Confidence Level(95.0%) = 13.96 + 2.07 = 16.03

Hence confidence interval given in option C is correct.