Question

Late bloomers? Japanese cherry trees tend to blossom early when spring weather is warm and later when spring weather is cool. Here are some data on the average March temperature (in degrees Celsius) and the day in April when the first cherry blossom appeared over a 24-year period:

a. Make a well-labeled scatterplot that’s suitable for predicting when the cherry trees will blossom from the temperature. Which variable did you choose as the explanatory variable? Explain your reasoning.

b. Use technology to calculate the correlation and the equation of the least-squares regression line. Interpret the slope and y-intercept of the line in this setting.

c. Suppose that the average March temperature this year was 8.2°C. Would you be willing to use the equation in part (b) to predict the date of the first blossom? Explain your reasoning.

d. Calculate and interpret the residual for the year when the average March temperature was 4.5°C.

e. Use technology to help construct a residual plot. Describe what you see.

Short answer

Part (b) y= 33.1203 − 4.6855x

Part (c) No.

Part (d) Residual is −2.033

Part (a)

Part (e) The linear regression line seems to be a good fit.

Step-by-step solution

Part (a) Step 1: Given information

Part (a) Step 2: Explanation

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:

Part (b) Step 1: Calculation

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:

$$\begin{gathered} \stackrel{⏜}{y}=a+bx \\ \stackrel{⏜}{\Rightarrow y}=33.1203-4.6855x \end{gathered}$$

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 $0°C$

Part (c) Step 1: Calculation

The regression line in part (b) is:

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

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

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).

Part (d) Step 1: Explanation

The regression line in part (b) is:

$\stackrel{⏜}{y}=33.1203-4.6855x$

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

is:

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

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

Thus, the residual is as:

$$\begin{gathered} Residual=y-\stackrel{⏜}{y} \\ =10-12.033 \\ =-2.033 \end{gathered}$$

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.

Part (e) Step 1: Explanation

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.