The data show the time intervals ater an eruption (o the next eruption) of a cer

Question

The data show the time intervals ater an eruption (o the next eruption) of a certain geyser Find the regression equation, lettling the fest variabile be the independent (a) variable Find the best predicted Sime of the inderval ater an ersption gverthat ament eruption has a height of 105 foet Use a sig lance level of 0.05 115 133 137140 99 112 100 126 Interval after l Cick the icon to view the critical values of the Pearson conelation coefficient r What is the regression equation? (Round to two decimal places as needed) What is the best predicted sime for the interval after an eruption that is 105 feet hight? ysLminutes(Rond to one decimal place as oeeded )

Explanation / Answer

Solution:

Confirm with data plese:

height <- c(115,133,137,140,99,112,100,126)
interval <- c(75,85,94,85,66,88,68,82)

Run the regressioj in R:

If the given data is above one:

Rcode :height <- c(115,133,137,140,99,112,100,126)
interval <- c(75,85,94,85,66,88,68,82)
lmod1 <- lm(interval~height)
summary(lmod1)

Output:

summary(lmod1)

Call:

lm(formula = interval ~ height)

Residuals:

Min 1Q Median 3Q Max

-5.2229 -3.0127 -2.0051 0.3867 11.7387

Coefficients:

Estimate Std. Error t value Pr(>|t|)  

(Intercept) 20.4149 17.3603 1.176 0.2842  

height 0.4986 0.1432 3.481 0.0131 *

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 6.117 on 6 degrees of freedom

Multiple R-squared: 0.6688, Adjusted R-squared: 0.6136

F-statistic: 12.12 on 1 and 6 DF, p-value: 0.01313

Regression eq is

y^=20.41+0.50(height)

ANSWER:

y^=20.41+0.50(height)

Solutionb;

for height =105

substitute in Reg eq

y^=20.41+0.50(height)

y^=20.41+0.50(105)

y^=72.91

y^=72.9

ANSWER 72.9

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