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 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 45 landscapers and 31 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 47.363 2 58.492 3 48.700 4 51.115 5 52.224 6 48.547 7 49.258 8 49.289 9 42.657 10 47.655 11 49.494 12 48.440 13 46.669 14 51.322 15 42.519 16 42.797 17 41.541 18 39.540 19 45.505 20 47.675 21 52.216 22 45.817 23 48.901 24 45.650 25 49.115 26 52.551 27 43.138 28 52.703 29 52.066 30 52.145 31 53.698 32 48.133 33 48.788 34 48.111 35 45.038 36 42.649 37 50.008 38 50.163 39 55.597 40 45.957 41 54.454 42 44.858 43 52.324 44 48.494 45 47.537

Table 2. Office Workers Subject Vitamin D 1 40.445 2 40.820 3 50.853 4 53.302 5 53.123 6 50.409 7 46.463 8 52.296 9 52.439 10 50.728 11 52.007 12 48.892 13 53.052 14 51.037 15 49.433 16 56.822 17 49.609 18 53.978 19 48.579 20 45.722 21 46.759 22 52.477 23 50.545 24 45.033 25 48.982 26 54.726 27 48.743 28 43.796 29 47.321 30 54.831 31 45.392

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. -2.63 to 0.75 ng/mL

b. -3.30 to 0.61 ng/mL

c. -3.46 to 0.59 ng/mL

d. -2.99 to 0.65 ng/mL

Answer 1

Sample #1   ----> 1
mean of sample 1,    x̅1=   48.465
standard deviation of sample 1,   s1 =    3.9746
size of sample 1,    n1=   45
      
Sample #2   ----> 2
mean of sample 2,    x̅2=   49.633
standard deviation of sample 2,   s2 =    3.9699
size of sample 2,    n2=   31

α=0.05

Degree of freedom, DF=   n1+n2-2 =    74              
t-critical value =    t α/2 =    1.9925   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.9727              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.9273              
margin of error, E = t*SE =    1.9925   *   0.93   =   1.84762  
                      
difference of means =    x̅1-x̅2 =    48.4647   -   49.633   =   -1.1680
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -1.1680   -   1.8476   =   -3.016
Interval Upper Limit=   (x̅1-x̅2) + E =    -1.1680   +   1.8476   =   0.680

so, answer is

d. -2.99 to 0.65 ng/mL