The regression equation is
temperature = 35.78 + 0.2512 chirps/min
S = 1.21526 R-Sq = 95.7% R-Sq(adj) = 95.5%
Analysis of Variance
Source DF SS MS F P
Regression 1 914.570 914.570 619.27 0.000
Error 28 41.352 1.477
Total 29 955.922
Given that Sxx= 14,497.9 and\(\style{font-family:Times New Roman;}{\overline x=}\)117.067, what is a 95% confidence interval for the mean air temperature when a cricket is chirping at a frequency of 120 chirps per minute?
(66.539, 67.301)
(63.818, 68.022)
(65.462, 66.379)
(62.389, 67.451)
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