91Ó°ÊÓ

Temperature is the explanatory variable, and the days in April to first blossom are the response variable, because we expect temperature to influence the days in April to first blossom. As a result, the scatterplot looks like this:

Select $1$: Edit using a calculator by pressing $STAT$Then, in the list$L_1$put the sugar data, and in the list $L_2$enter the calorie data.

Next, press on $STAT$select $CALC$and then select $Linreg(a+bx)$Next we need to finish the command by entering $L_1{L}_2$

$Linreg(a+bx)L_1{L}_2$

Finally, pressing on entering then gives us the following result:

$$\begin{gathered} y=a+bx \\ a=33.1203 \\ b=−4.6855 \\ r=−0.8511 \end{gathered}$$

This then implies the regression line as:

$$\begin{gathered} \stackrel{Áåœ}{y}=a+bx \\ \stackrel{Áåœ}{⇒y}=33.1203−4.6855x \end{gathered}$$

As a result, the days from the first flower in April fall by $4.6855$ days per degree Celsius on average. And the days from the first flower in April are $33.1203$ days when the temperature is $0°C$

The regression line in part (b) is:

To predict the date to first blossom at $4.2C$is then,

$$\begin{gathered} \stackrel{Áåœ}{y}=33.1203−4.6855x \\ =33.1203−4.6855(4.2) \\ =−5.3052 \end{gathered}$$

We then see that the first flower is expected in April at $5.3052$ However, because a day is always a positive integer, this does not make it dense, therefore we are unwilling to utilize the equation in part (b).

The regression line in part (b) is:

$\stackrel{Áåœ}{y}=33.1203−4.6855x$

Now, the days to first blossom when average March temperature was $4.5C$

is:

$$\begin{gathered} \stackrel{Áåœ}{y}=33.1203−4.6855x \\ =33.1203−4.6855(4.5) \\ =12.033 \end{gathered}$$

And the actual value is $10$from the table given.

Thus, the residual is as:

$$\begin{gathered} Residual=y−\stackrel{Áåœ}{y} \\ =10−12.033 \\ =−2.033 \end{gathered}$$

This means that while using the regression line with temperature as the explanatory variable, we overestimated the number of days in April till the first flower by $2.033$ days.

The residual plot is as:

As a result, the residual plot shows no discernible pattern, and the linear regression line appears to be a decent fit.