Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Answer:
The only one that makes sense IF the model behaves as the Earth is D.
Explanation:
If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.
We need the reading for this I think
