Temperature (In °F) # of Adults 96.0-96.4 2 96.5-96.9 4 97.0-97.4 13 97.5-97.9 21 98.0-98.4 38 98.5-98.9 33 99.0-99.4 15 99.5-99.9 100.0-100.4 1 100.5-100.9 1 “Data for the JAMA article were collected from healthy men and women, ages 18 to 40, at the University of Maryland Center for Vaccine Development, Baltimore. Note: The temperature given is their “normal” body temperature. 1. (5 pts) How many adults did Allen Shoemaker sample in his study on “normal” body temperature? Note: The sample size from #1 should remain the sample size in your calculator (for all TESTS) for the rest of the lab! 2. (5 pts) Enter the midpoints in Li and the frequencies in La, and use 1-VarStats to calculate the mean of the frequency distribution (see section 3.2 if needed) The sample mean of Shoemaker data: (2 decimal places)
The Correct Answer and Explanation is :
Question 1: Total number of adults sampled
To find how many adults Allen Shoemaker sampled in his study, we need to sum the frequencies provided in the table for each temperature range.
Here are the frequencies listed:
- 96.0–96.4: 2
- 96.5–96.9: 4
- 97.0–97.4: 13
- 97.5–97.9: 21
- 98.0–98.4: 38
- 98.5–98.9: 33
- 99.0–99.4: 15
- 99.5–99.9: 1
- 100.0–100.4: 1
- 100.5–100.9: 1
Now, we sum all these frequencies:
[
2 + 4 + 13 + 21 + 38 + 33 + 15 + 1 + 1 + 1 = 129
]
Thus, 129 adults were sampled in Shoemaker’s study.
Question 2: Calculate the sample mean
To calculate the sample mean of the frequency distribution, we need to use the formula for the mean of grouped data:
[
\text{Mean} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}
]
Where:
- ( f_i ) is the frequency for each group
- ( x_i ) is the midpoint for each group
Step 1: Calculate midpoints ( x_i )
The midpoint of each temperature range is calculated by averaging the lower and upper bounds of each range:
- 96.0–96.4: Midpoint = ( \frac{96.0 + 96.4}{2} = 96.2 )
- 96.5–96.9: Midpoint = ( \frac{96.5 + 96.9}{2} = 96.7 )
- 97.0–97.4: Midpoint = ( \frac{97.0 + 97.4}{2} = 97.2 )
- 97.5–97.9: Midpoint = ( \frac{97.5 + 97.9}{2} = 97.7 )
- 98.0–98.4: Midpoint = ( \frac{98.0 + 98.4}{2} = 98.2 )
- 98.5–98.9: Midpoint = ( \frac{98.5 + 98.9}{2} = 98.7 )
- 99.0–99.4: Midpoint = ( \frac{99.0 + 99.4}{2} = 99.2 )
- 99.5–99.9: Midpoint = ( \frac{99.5 + 99.9}{2} = 99.7 )
- 100.0–100.4: Midpoint = ( \frac{100.0 + 100.4}{2} = 100.2 )
- 100.5–100.9: Midpoint = ( \frac{100.5 + 100.9}{2} = 100.7 )
Step 2: Calculate ( f_i \cdot x_i ) for each group
Now, we multiply each frequency ( f_i ) by its corresponding midpoint ( x_i ):
- ( 2 \times 96.2 = 192.4 )
- ( 4 \times 96.7 = 386.8 )
- ( 13 \times 97.2 = 1263.6 )
- ( 21 \times 97.7 = 2051.7 )
- ( 38 \times 98.2 = 3731.6 )
- ( 33 \times 98.7 = 3257.1 )
- ( 15 \times 99.2 = 1488.0 )
- ( 1 \times 99.7 = 99.7 )
- ( 1 \times 100.2 = 100.2 )
- ( 1 \times 100.7 = 100.7 )
Step 3: Sum all ( f_i \cdot x_i ) values
[
192.4 + 386.8 + 1263.6 + 2051.7 + 3731.6 + 3257.1 + 1488.0 + 99.7 + 100.2 + 100.7 = 11671.7
]
Step 4: Calculate the mean
Now, use the formula for the mean:
[
\text{Mean} = \frac{11671.7}{129} \approx 90.5
]
Thus, the sample mean of Shoemaker’s data is 90.5°F.
Conclusion:
- Allen Shoemaker sampled 129 adults.
- The mean of the frequency distribution for their normal body temperature is 90.5°F.