Thank you for visiting Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM Temperature ∠F at 8 AM 98. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Mean difference (d¯) = 0.32 and standard deviation (sd) ≈ 0.3036.
To find the mean difference (d¯) and standard deviation (sd) of the paired data.
Calculate the differences between each pair of temperatures (12 AM - 8 AM).
Find the mean and standard deviation of these differences.
The differences for each pair:
1. 99.2 - 98.5 = 0.7
2. 99.8 - 99.2 = 0.6
3. 97.7 - 97.4 = 0.3
4. 97.8 - 97.9 = -0.1
5. 97.5 - 97.4 = 0.1
Mean difference (d¯) = (0.7 + 0.6 + 0.3 - 0.1 + 0.1) / 5
Mean difference (d¯) = 1.6 / 5
Mean difference (d¯) = 0.32
Calculate the squared differences from the mean difference:
(0.7 - 0.32)² = 0.1449
(0.6 - 0.32)² = 0.0784
(0.3 - 0.32)² = 0.0009
(-0.1 - 0.32)² = 0.1849
(0.1 - 0.32)² = 0.0516
Find the average of these squared differences:
Average squared difference = (0.1449 + 0.0784 + 0.0009 + 0.1849 + 0.0516) / 5 = 0.09214
Take the square root of the average squared difference to find the standard deviation (sd):
sd = √0.09214
sd ≈ 0.3036
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