Question

The data show the time intervals after an eruption​ (to the next​ eruption) of a certain geyser. Find the regression​ equation, letting the height of the current eruption be the explanatory variable​ (denoted by​ x). Then use this equation to determine the predicted length of the time interval after an eruption given that the current eruption has a height of 113feet.

Height (ft), Interval after (min)

96 66

128 85

75 59

128 86

88 70

73 75

80 73

96 75

What is the regression​ equation?

y = _____________+______________x​ (Round to three decimal places as​ needed.

What is the predicted time of the​ interval?

y ≈ _____________minutes(Round to one decimal place as​ needed.

Answer 1

The statistical software output for this problem is:

Simple linear regression results:
Dependent Variable: Interval
Independent Variable: Height
Interval = 42.865183 + 0.32209233 Height
Sample size: 8
R (correlation coefficient) = 0.77793416
R-sq = 0.60518156
Estimate of error standard deviation: 6.1344219

Parameter estimates:

Parameter	Estimate	Std. Err.	Alternative	DF	T-Stat	P-value
Intercept	42.865183	10.372232	≠ 0	6	4.1326866	0.0061
Slope	0.32209233	0.10620883	≠ 0	6	3.0326323	0.023

Analysis of variance table for regression model:

Source	DF	SS	MS	F-stat	P-value
Model	1	346.0882	346.0882	9.1968586	0.023
Error	6	225.7868	37.631133
Total	7	571.875

Predicted values:

X value	Pred. Y	s.e.(Pred. y)	95% C.I. for mean	95% P.I. for new
113	79.261616	2.8563067	(72.272485, 86.250746)	(62.703846, 95.819385)

Hence,

Regression equation:

y = 42.865 + 0.322 x

Prediction time of the interval

y = 79.3 minutes