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ttools ttools
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A year ago
In the late 1890s, scientists first noted that the frequency of a cricket’s chirps is related to air temperature. For each of 30 crickets, the frequency of chirping (in chirps per minute) and the air temperature (in degrees Fahrenheit) were recorded, and a regression analysis was performed.

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\(\style{font-family:Times New Roman;}{\overline x=}\)067, what is a 90% prediction interval for the 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)

Textbook 
Introductory Statistics: A Problem-Solving Approach

Introductory Statistics: A Problem-Solving Approach


Edition: 3rd
Author:
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ftricey04ftricey04
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A year ago
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ttools Author
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A year ago
Thank you, thank you, thank you!
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Yesterday
This helped my grade so much Perfect
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2 hours ago
Good timing, thanks!
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