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Recall in our discussion of the normal distribution the research study that examined the blood vitamin...

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 collect data from US office workers to examine the difference between the average vitamin D levels of landscapers and office workers, which will reflect any occupational sun exposure differences as measured by blood vitamin D levels. You obtain research funding to sample at random 32 landscapers and 32 office workers, collect blood samples, and send these samples to your collaborating lab in order to quantify the amount of vitamin D in both groups' 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 tables.

Table 1. Landscapers
Subject Vitamin D
1 43.104
2 45.147
3 43.588
4 46.764
5 43.083
6 50.448
7 39.430
8 48.074
9 40.328
10 44.983
11 39.339
12 39.977
13 43.447
14 40.712
15 51.308
16 42.451
17 42.774
18 54.254
19 44.717
20 40.029
21 31.471
22 38.238
23 47.022
24 41.168
25 35.647
26 44.090
27 45.703
28 52.699
29 46.258
30 46.365
31 53.494
32 44.693
Table 2. Office Workers
Subject Vitamin D
1 28.820
2 29.225
3 36.528
4 33.125
5 33.972
6 41.400
7 32.249
8 32.337
9 32.130
10 28.187
11 40.917
12 34.138
13 38.358
14 30.696
15 32.616
16 30.722
17 33.495
18 32.288
19 33.601
20 30.804
21 35.609
22 34.327
23 28.659
24 36.964
25 40.766
26 37.309
27 34.924
28 36.628
29 35.763
30 29.431
31 34.400
32 23.855

What is the estimated 95% confidence interval (CI) of the average difference in blood vitamin D levels between US landscapers and office workers in ng/mL? Assign groups 1 and 2 to be landscapers and office workers, respectively.

Please note the following: 1) in practice, you as the analyst decide how to assign groups 1 and 2 and subsequently interpret the results appropriately in the context of the data, though for the purposes of this exercise the groups are assigned for you; 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. 7.47 to 14.01 ng/mL

b. 7.43 to 14.71 ng/mL

c. 8.30 to 12.74 ng/mL

d. 6.97 to 10.95 ng/mL

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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 collect data from US office workers to examine the difference between the average vitamin D levels of landscapers and office workers, which will reflect any occupational sun exposure differences as measured by blood vitamin D levels. You obtain research funding to sample at random 32 landscapers and 12 office workers, collect blood samples, and send these samples to your collaborating lab in order to quantify the amount of vitamin D in both groups' 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 tables.

Table 1. Landscapers
Subject Vitamin D
1 36.225
2 33.841
3 45.296
4 33.166
5 38.312
6 38.569
7 50.759
8 35.712
9 32.403
10 47.064
11 34.893
12 33.346
13 40.035
14 40.545
15 32.232
16 40.780
17 20.102
18 48.560
19 40.945
20 37.296
21 39.898
22 36.296
23 49.822
24 41.920
25 43.352
26 31.312
27 33.932
28 34.469
29 32.243
30 35.386
31 44.257
32 33.247
Table 2. Office Workers
Subject Vitamin D
1 49.436
2 44.957
3 43.747
4 56.000
5 46.957
6 54.940
7 49.879
8 46.987
9 52.289
10 47.720
11 47.480
12 57.264

What is the estimated 95% confidence interval (CI) of the average difference in blood vitamin D levels between US landscapers and office workers in ng/mL? Assign groups 1 and 2 to be landscapers and office workers, respectively.

Please note the following: 1) in practice, you as the analyst decide how to assign groups 1 and 2 and subsequently interpret the results appropriately in the context of the data, though for the purposes of this exercise the groups are assigned for you; 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. -17.92 to -6.50 ng/mL

b. -17.04 to -8.24 ng/mL

c. -15.85 to -7.74 ng/mL

d. -18.23 to -6.81 ng/mL

Homework Answers

Answer #1

1)

sample mean 'x̄= 10.518
sample size    n= 32.00
std deviation σ= 6.78
std error ='σx=σ/√n= 1.1980
for 95 % CI value of z= 1.96
margin of error E=z*std error = 2.35
lower bound=sample mean-E= 8.20
Upper bound=sample mean+E= 12.74

Correct option is c: 8.30 to 12.74

2)

Pooled Std dev Sp=√((n1-1)s21+(n2-1)*s22)/(n1+n2-2)= 5.9389
Point estimate : x1-x2= -11.7979
standard error se =Sp*√(1/n1+1/n2)= 2.010
point estimate of difference=x1-x2= -11.798
for 95 % CI & 42 df value of t= 2.018
margin of error E=t*std error = 4.057
lower bound=mean difference-E= -15.855
Upper bound=mean differnce +E= -7.741
from above 95% confidence interval for population mean =(-15.85 , -7.74)

option C is correct c. -15.85 to -7.74 ng/mL

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