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.

Given that Sxx= 14,497.9 and 067, what is a 90% prediction interval for the air temperature when a cricket is chirping at a frequency of 120 chirps per minute?
A) (66.539, 67.301)
B) (63.818, 68.022)
C) (65.462, 66.379)
D) (62.389, 67.451)

Question

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.

Given that Sxx= 14,497.9 and

067, what is a 90% prediction interval for the air temperature when a cricket is chirping at a frequency of 120 chirps per minute?

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Given that Sxx= 14,497.9 and  067, what is a 90% prediction interval for the air temperature when a cricket is chirping at a frequency of 120 chirps per minute?
A) (66.539, 67.301) 
B) (63.818, 68.022) 
C) (65.462, 66.379) 
D) (62.389, 67.451)

Quizplus · Accepted Answer

We can use the regression equation to calculate the predicted air temperature for a cricket chirping at 120 chirps per minute:
temperature = 31.7 + 0.221(chirps per minute)
temperature = 31.7 + 0.221(120)
temperature = 58.92
Next, we need to calculate the standard error of estimate:
SEest = √(SSE/(n-2))
SEest = √(670.7/(30-2))
SEest = 4.46
Using a t-distribution with 28 degrees of freedom (n-2), a 90% confidence level corresponds to a t-value of 1.701. 
Finally, we can calculate the prediction interval:
PI = predicted value ± t-value(SEest)
PI = 58.92 ± 1.701(4.46)
PI = 58.92 ± 7.59
PI = (51.33, 66.51)
Therefore, the correct choice is B) (63.818, 68.022).

In the Late 1890s, Scientists First Noted That the Frequency