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 44 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 41.660 36.567 2 31.816 33.510 3 33.769 37.652 4 38.861 39.134 5 34.790 30.753 6 36.213 35.072 7 44.393 32.603 8 35.554 35.049 9 31.441 36.983 10 38.545 39.562 11 33.054 33.609 12 32.937 37.888 13 35.697 40.524 14 29.110 40.090 15 38.639 30.982 16 39.473 32.265 17 32.730 40.592 18 32.579 40.416 19 32.652 44.457 20 45.973 37.926 21 36.431 32.522 22 33.184 34.315 23 45.950 40.684 24 41.087 37.266 25 34.263 35.425 26 31.785 36.698 27 31.899 39.583 28 29.466 39.082 29 44.550 32.102 30 34.123 37.685 31 37.132 41.292 32 37.199 38.958 33 29.741 31.425 34 37.353 41.583 35 32.760 44.935 36 29.158 41.590 37 38.875 41.280 38 44.192 34.898 39 33.254 33.277 40 30.917 35.957 41 40.127 38.090 42 33.147 34.123 43 41.151 32.412 44 36.595 32.143 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. -1.02 to 2.78 ng/mL b. -1.12 to 3.19 ng/mL c. -1.17 to 3.12 ng/mL d. -1.13 to 2.53 ng/mL

Answer 1

For calculating confidence interval first we need to find mean of difference and standard deviation of difference.

20 21 45.973 36.431 33.184 45.950 41.087 34.263 31.785 31.899 29.466 44.550 34.123 37.132 37.199 29.741 37.353 32.760 29.158 38.875 44.192 33.254 37.926 32.522 34.315 40.684 37.266 35.425 36.698 39.583 39.082 32.102 37.685 41.292 38.958 31.425 41.583 44.935 41.590 41.280 34.898 33.277 23 24 25 26 27 28 29 30 31 32 8.047-8.9273 79.6967 3.909 4.7893 22.9374 .1310.2507 0.0629 5.266-6.1463 37.7770 .821-4.7013| 22.1022 1.1620.2817 0.0794 4.9134.0327 16.2627 7.684 6.803746.2903 9.6168.7357 76.3125 12.448-13.3283 177.6436 34 35 36 37 38 39 3.562 2.6817 7.1915 4.160 3.2797 10.7564 1.7590.8787 0.7721 1.6840.8037 0.6459 4.230 3.3497 11.2205 12.175 11.2947 127.5702 12.43211.5517 133.4418 2.4051.5247 2.3247 9.294 -10.1743 103.5164 0.023 0.8573 0.7350

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

Step 2: Find t_?/2
Calculate t_?/2 by using t-distribution with degrees of freedom (DF) as n - 1 = 44 - 1 = 43 and ?/2 = 0.025 as right-tailed area and left-tailed area.

t_?/2 = 2.016692 (Obtained using t value calculator screenshot attached)

Step 3: Calculate Confidence Interval

Confidence Formula: [? - t_?/2•(s_d/?n) , ? + t_?/2•(s_d/?n)]

Here ? = 0.8803, s_d = 6.4183, t_?/2 = 2.0166692
Lower Bound = ? - t_?/2•(s_d/?n) = 0.8803 - (2.01666)(6.42/?44) = -1.07
Upper Bound = ? + t_?/2•(s_d/?n) = 0.8803 + (2.01666)(6.42/?44) = 2.83

Confidence Interval = (-1.07, 2.83)

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