Question

In: Statistics and Probability

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 38 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 60.474 2 61.879 3 58.536 4 60.255 5 53.001 6 56.432 7 51.904 8 50.804 9 57.783 10 45.492 11 50.540 12 51.947 13 54.618 14 51.663 15 52.344 16 44.544 17 54.957 18 54.532 19 48.846 20 62.853 21 48.515 22 52.759 23 55.749 24 50.077 25 50.321 26 58.301 27 47.651 28 51.767 29 46.071 30 49.933 31 50.437 32 52.420 Table 2. Office Workers Subject Vitamin D 1 31.488 2 21.113 3 44.218 4 37.891 5 31.980 6 35.140 7 36.631 8 33.028 9 37.242 10 32.571 11 30.206 12 29.767 13 23.423 14 43.800 15 28.851 16 34.948 17 34.758 18 45.972 19 35.608 20 34.985 21 33.014 22 40.407 23 33.622 24 33.221 25 38.673 26 32.804 27 39.355 28 37.076 29 41.068 30 32.902 31 36.327 32 28.681 33 36.043 34 37.275 35 39.641 36 38.616 37 38.252 38 28.623 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. 15.72 to 20.41 ng/mL b. 14.62 to 22.45 ng/mL c. 17.22 to 23.57 ng/mL d. 13.99 to 22.65 ng/mL

Solutions

Expert Solution

For calculating confidence interval first we need to find mean and standard deviation for both the groups

We are calculating confidence interval based on assumption that population standard deviation are not equal

x?? (sample mean 1) = 53.0439 x?? (sample mean 2) = 34.9795  
s? (sample standard deviation 1) = 4.7007 s? (sample standard deviation 2) = 5.1999
n? (sample size 1) = 32 n? (sample size 2) = 38

Step 1: Find ?/2
Level of Confidence = 95%
? = 100% - (Level of Confidence) = 5%
?/2 = 2.5% = 0.025

Step 2: Find degrees of freedom and t?/2
Degrees of freedom = smaller of (n1 - 1 , n2 - 1 ) = smaller of (31 , 37) = 31
Calculate t?/2 by using t-distribution with degrees of freedom (DF) = 31 and ?/2 = 0.025 as right-tailed area and left-tailed area.

t?/2 = 2.0395 (Obtained using online t value calculator screenshot attached)

Step 3: Calculate Confidence Interval

t?/2 = 2.0395
Standard Error = ? (s?)²/n? + (s?)²/n? = ?1.4020697195230265 = 1.1841
Lower Bound = (x?? - x??) - t?/2•(? (s?)²/n? + (s?)²/n? ) = (53.0439 - 34.9795) - (2.0395)(1.1841) = 15.65
Upper Bound = (x?? + x??) + t?/2•(? (s?)²/n? + (s?)²/n? ) = (53.0439 - 34.9795) + (2.0395)(1.1841) = 20.48
Confidence Interval = (15.65, 20.48)

Option a is correct as confidence interval is closest to it.


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