Question

The data show the bug chirps per minute at different temperatures. Find the regression​ equation, letting the first variable be the independent​ (x) variable. Find the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute. Use a significance level of 0.05. What is wrong with this predicted​ value?

Chirps in 1 min	758	1226	1021	1221	1182	1178
Temperature ​(degrees°​F)	65.1	92.8	84.3	94.9	85.5	91.2

What is the regression​ equation?

ModifyingAbove y with caretyequals=nothingplus+nothingx

​(Round the​ x-coefficient to four decimal places as needed. Round the constant to two decimal places as​ needed.)

Answer 1

We can run the regression in Excel by going to Data -> Data analysis -> Regression. The output is:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.964764
R Square	0.930769
Adjusted R Square	0.913461
Standard Error	3.199979
Observations	6
ANOVA
	df	SS	MS	F	Significance F
Regression	1	550.6739	550.6739	53.77746	0.001841
Residual	4	40.95946	10.23986
Total	5	591.6333
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	22.50649	8.706796	2.584934	0.061009	-1.66745	46.68043	-1.66745	46.68043
Chirps in 1 min	0.05751	0.007842	7.333311	0.001841	0.035736	0.079284	0.035736	0.079284

We can see that the regression equation is as follows (from the table):

y = 22.5065 + 0.0575x

where y = Dependent variable = Temperature (F)

x = Chirps in 1 min

Best predicted temperature when x = 3000,

y = 22.5065 + 0.0575*3000 = 195.0065

This predicted value is wrong as this temperature isn't possible in reality and it is a hypothetical situation when the chirping is at such a high rate of 3000 per minute.