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1. Explain the regression model and the regression model assumptions.
2. Perform the hypothesis test for the slope ${\beta }_1$ of the population regression equation.
3. Construct confidence intervals for the slope ${\beta }_1$.
4. Use confidence intervals to perform the hypothesis test for the slope ${\beta }_1$.

1. Construct and interpret a scatterplot of the response variable $y$ versus the predictor variable $x$.
2. Calculate and interpret the correlation coefficient $r$.
3. Compute the regression equation $\hat{y}=b_{1}x+b_0$. Interpret the meaning of the $y$ intercept $b_0$ and the slope $b_1$ of the regression equation.
4. Predict the number of text messages sent by a 20-year-old person, and calculate the prediction error (residual).

1. The number of messages depends on age, and not vice versa, so the predictor variable $x$ is age and the response variable $y$ is messages. Also, note that in (d) we are trying to predict the number of text messages, which tells us that messages is the response variable $y$ because we never try to predict the known value of $x$. The TI-83/84 scatterplot is shown in Figure 1. As age increases, the number of messages tends to decrease.
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   FIGURE 1 TI-83/84 scatterplot of messages versus age.
2. Figure 2 shows the correlation coefficient $r\approx -0.9701$, calculated by the TI-83/84. Age and messages are negatively correlated. An increase in age is associated with a decrease in the number of messages.

3. Figure 2 shows that $a=b_1=-1.5$ and $b=b_0=60.6$, and thus the regression equation is

   $\hat{y}=b_{1}x+b_0=\left(-1.5\right)\left(\text{age}\right)+60.6$

   We can interpret $b_0$ and $b_1$ as follows:

   • The $y$ intercept $b_0=60.6$ is the estimated number of text messages sent by someone aged $x=0$, which does not make sense because this value $x=0$ lies far below the minimum value of $x$ and therefore represents extreme extrapolation.
   • The slope $b_1=-1.5$ means there is an estimated decrease of 1.5 in the number of text messages for each additional year of age.
   

   FIGURE 2 TI-83/84 correlation and regression results.

4. For a 20-year-old person, the estimated number of daily text messages is

   $\hat{y}=b_{1}x+b_0=\left(-1.5\right)\left(20\right)+60.6=30.6$

   The actual number of text messages sent by our 20-year-old in Table 1 is $y=29$. Our prediction from (c) is $\hat{y}=30.6$. Thus, our prediction error (or residual) is: $\left(y-\hat{y}\right)=\left(29-30.6\right)=-1.6$. Our 20-year-old sent slightly fewer text messages than expected.

YOUR TURN#1

1. Construct and interpret a scatterplot of the response variable $y$ versus the predictor variable $x$.
2. Calculate and interpret the correlation coefficient $r$.
3. Compute the regression equation $\hat{y}=b_{1}x+b_0$. Interpret the meaning of the $y$ intercept $b_0$ and the slope $b_1$ of the regression equation.
4. Predict the score for a 22-year old person, and calculate the prediction error (residual).

Note that there is no “hat” on the $y$ in the population regression equation because the equation represents a model of the relationship between the actual values of $x$ and $y$, not an estimate of $y$.

$y={\beta }_{1}x+{\beta }_0+\epsilon$

1. Zero-mean assumption. The error term $\epsilon$ is a random variable, with a mean of 0. That is, the expected value of the random variable $\epsilon$ is $0:E\left(\epsilon \right)=0$.
2. Constant variance assumption. The variance of $\epsilon$, which is denoted as ${\sigma }^2$, is the same regardless of the value of $x$.
3. Independence assumption. The values of $\epsilon$ are independent of each other.
4. Normality assumption. The error term $\epsilon$ is a normal random variable.

Developing Your Statistical Sense

1. Calculate the residuals $y-\hat{y}$.
2. Verify the regression assumptions.

1. Table 2 contains the $x$ and $y$ data from Table 1, the fitted (predicted) values $\hat{y}$, and the residuals $y-\hat{y}$.
   Table 13.3: Table 2 Calculating the residuals

   $x=\text{Age}$	$y=\text{Text messages}$	Fitted (predicted) values $\hat{y}=\left(-1.5\right)\left(\text{age}\right)+60.6$	Residuals $y-\hat{y}$
   18	35	33.6	1.4
   20	29	30.6	−1.6
   22	27	27.6	−0.6
   24	28	24.6	3.4
   26	19	21.6	−2.6
   28	16	18.6	−2.6
   30	19	15.6	3.4
   32	12	12.6	−0.6
   34	8	9.6	−1.6
   36	8	6.6	1.4

2. The scatterplot in Figure 5 of the residuals versus fitted values shows no strong evidence of the unhealthy patterns shown in Figure 4. Thus, the independence assumption, the constant variance assumption, and the zero-mean assumption are verified. Also, the normal probability plot of the residuals in Figure 6 indicates no evidence of departures from normality in the residuals. Therefore, we conclude that the regression assumptions are verified.
   

   FIGURE 5 Scatterplot of residuals versus fitted values.
   

   FIGURE 6 Normal probability plot of the residuals.

YOUR TURN#2

1. Calculate the residuals $y-\hat{y}$. for the regression of score on age.
2. Verify the regression assumptions.

$t_{\text{data}}={\displaystyle \frac{b_1-{\beta }_1}{s/\sqrt{\sum {\left(x-\overline{x}\right)}^2}}}={\displaystyle \frac{b_1-0}{s/\sqrt{\sum {\left(x-\overline{x}\right)}^2}}}={\displaystyle \frac{b_1}{s/\sqrt{\sum {\left(x-\overline{x}\right)}^2}}}$

Here, $s$ refers to the standard error of the estimate, not the sample standard deviation.

1. Find $b_1$, the slope of the regression line.
2. Calculate $s$, the standard error of the estimate.
3. Compute $\sqrt{\sum {\left(x-\overline{x}\right)}^2}$, the numerator of the sample variance of the $x$ data.

1. From Example 1, the slope of the regression line is $b_1=-1.5$.

2. Recall from Section 4.3 (page 228) that

   $s=\sqrt{{\displaystyle \frac{\text{SSE}}{n-2}}}=\sqrt{{\displaystyle \frac{\sum {\left(y-\hat{y}\right)}^2}{n-2}}}=\sqrt{{\displaystyle \frac{\sum {\left(\text{residual}\right)}^2}{n-2}}}$

   is the standard error of the estimate. Squaring each residual from Table 2 gives us the squared residuals in Table 3, and the sum of squared residuals, or sum of squares error, equal to

   $\text{SSE}={\sum \left(y-\hat{y}\right)}^2=46.4$

   Then the standard error of the estimate is

   $s=\sqrt{{\displaystyle \frac{\text{SSE}}{n-2}}}=\sqrt{{\displaystyle \frac{46.4}{8}}}\approx \mathrm{2.408318916.}$

   Table 13.4: Table 3 Calculating SSE

   Residuals $y-\hat{y}$	Squared residuals ${\left(y-\hat{y}\right)}^2$
   1.4	1.96
   −1.6	2.56
   −0.6	0.36
   3.4	11.56
   −2.6	6.76
   −2.6	6.76
   3.4	11.56
   −0.6	0.36
   −1.6	2.56
   1.4	1.96
   	$\text{Sum}=46.4$

3. To compute $\sum {\left(x-\overline{x}\right)}^2$ , we note from page 110 in Chapter 3 that the sample variance of $x$ is

   $s_x^2={\displaystyle \frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}}$

   Multiplying each side of the equation by $n-1$ , we obtain an equation for the quantity $\sum {\left(x-\overline{x}\right)}^2$:

   $\sum {\left(x-\overline{x}\right)}^2=\left(n-1\right)·s_x^2$

   The TI-83/84 output from Figure 7 shows that $s_x=6.055300708$ , and, because $n=10$,

   $\sum {\left(x-\overline{x}\right)}^2= \left(n-1\right)· s_x^2 = \left(9\right) {\left(6.055300708\right)}^2 = 330$

   Therefore,

   $t_{\text{data}} = {\displaystyle \frac{b_1}{s/\sqrt{\sum {\left(x-\overline{x}\right)}^2}}} = {\displaystyle \frac{- 1.5}{2.408318916 /\sqrt{330}}} \approx -11.3$

   

   FIGURE 7 Summary statistics for the $x$ (age) data.

• Step 1 State the hypotheses.
  • $H_0: {\beta }_1 = 0$ No linear relationship exists between age and text messages.
  • $H_a: {\beta }_1 \ne 0$ A linear relationship exists between age and text messages.

• Step 2 Find the $t$ critical value $t_{\text{crit}}$ and the rejection rule. To find $t_{\text{crit}}$, use the $t$ distribution table (Table D in the Appendix) for a two-tailed test and degrees of freedom $\text{df}=n-2$. The rejection rule for this two-tailed test is

  $\text{Reject} H_0 \text{if} t_{\text{data} }\ge t_{\text{crit}} \text{or} t_{\text{data}} \le - t_{\text{crit}}$

  Here, $n= 10$, so $\text{df}=8$. For level of significance $\alpha = 0.01$, the $t$ table gives us $t_{\text{crit} }= 3.355$. We will reject $H_0$ if $t_{\text{data} }\ge 3.335 \text{or} t_{\text{data}} \le - 3.335$.

• Step 3 Calculate $t_{\text{data}}$. From Example 3, we have

  $t_{data} = {\displaystyle \frac{b_1}{s/\sqrt{\sum {\left(x - \overline{x}\right)}^2}}} \approx - 11.3$

• Step 4 State the conclusion and the interpretation. Because $t_{\text{data} }\approx - 11.3 \le -3.335$, we reject $H_0$. There is evidence, at level of significance $\alpha = 0.01$, that ${\beta }_{1 } \ne 0$ and that a linear relationship exists between age and text messages.

• Step 1 State the hypotheses and the rejection rule.

  • $H_0 : {\beta }_1 = 0$ No linear relationship exists between time and score.
  • $H_a: {\beta }_{1 } \ne 0$ A linear relationship exists between time and score.
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  FIGURE 8 Residuals versus fitted values plot.
  

  FIGURE 9 Normal probability plot of the residuals.

  The rejection rule is: reject $H_0$ if the $p\text{-value}\le 0.01$.

• Step 2 Calculate $t_{\text{data}}$.

  $t_{\text{data} }= {\displaystyle \frac{b_1}{s/\sqrt{\sum {\left(x- \overline{x}\right)}^2}}}$

  From page 226 in Section 4.3, we have $b_1= 2$. From Example 13 in Chapter 4 on page 228, we have

  $s= \sqrt{{\displaystyle \frac{12}{8}}} \approx 1.224744871$

  From the TI-83/84 summary statistics, we have the standard deviation of the $x$ (time) data to be $s_x = 2.449489743$. Thus, using the relationship we learned in Example 3:

  $\sum {\left(x - \overline{x}\right)}^2= \left(\mathrm{n} - 1\right) · s_x^2 = \left(9\right) {2.449489743}^2 = 54$

  Therefore,

  $t_{\text{data} }= {\displaystyle \frac{b_1}{s/\sqrt{\sum {\left(x - \overline{x}\right)}^2}}} \approx {\displaystyle \frac{2}{1.224744871/\sqrt{54}}} = 12$

  

  TI-83/84 summary statistics for $x$ (time) data.

• Step 3 Find the $p$-value. For instructions, see the Step-by-Step Technology Guide on page 730. The regression results (including the $p$-value) for the TI-83/84, Excel, Minitab, and CrunchIt! are shown in Figures 10, 11, 12, and 13. (Differing results are due to rounding.)

  

  FIGURE 10 TI-83/84 regression results.
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  FIGURE 11 Excel regression result.
  

  FIGURE 12 Minitab regression results.
  

  FIGURE 13 Crunchit! regression results.

• Step 4 The $p$-value of about $0.000 \text{is} \le \alpha = 0.01$, so we reject $H_0$. Evidence exists, at level of significance $\alpha = 0.01$, for a linear relationship between time and score.

YOUR TURN#3

$b_1\pm t_{\alpha /2}·{\displaystyle \frac{s}{\sqrt{{\sum \left(x-\overline{x}\right)}^2}}}$

$E = t_{\alpha / 2}· {\displaystyle \frac{s}{\sqrt{\sum {\left(x - \overline{x}\right)}^2}}}$

• $b_{1 }= 2$,
• $s = 1.224744871$, and
• $\sum {\left(x - \overline{x}\right)}^{2 }= 54$.

$E = t_{\alpha /2}· {\displaystyle \frac{s}{\sqrt{\sum {\left(x - \overline{x}\right)}^2}}} = \left(2.306\right)\left({\displaystyle \frac{1.224744874}{\sqrt{54}}}\right) \approx 0.3843$

$b_1\pm E=2\pm 0.3843=\left(1.6157,2.3843\right)$

• The margin of error $E = 0.3843$ means that, when we repeatedly take samples from this population, most of the time the sample estimate $b_1$ will be within $E = 0.3843$ of the unknown value of the slope ${\beta }_1$ of the population regression line.
• We are 95% confident that the interval $\left(1.6157 , 2.3843\right)$ captures the slope ${\beta }_1$ of the population regression line.
• Because ${\beta }_1$ is the increase in memory-test score per added minute of memorization, we are 95% confident that, for each additional minute of memorization, the increase in memory-test score will lie between 1.6157 and 2.3843 points.

• If a $100\left(1-\alpha \right)\% t$ confidence interval for ${\beta }_1$ does not contain zero, then we would reject $H_0:{\beta }_1=0$ for level of significance $\alpha$, and conclude that a linear relationship exists between $x$ and $y$.
• If a $100\left(1-\alpha \right)\% t$ confidence interval for ${\beta }_1$ does contain zero, then we would not reject $H_0:{\beta }_1=0$ for level of significance $\alpha$.

1. Construct and interpret a 99% confidence interval for the slope ${\beta }_1$ for the text messaging data in Table 1.
2. Use the confidence interval in (a) to test whether a linear relationship exists between age and text messages, using level of significance $\alpha =0.01$.

1. The regression assumptions were verified in Example 2. Also,
   • In Example 1, we found $b_1=-1.5$.
   • In Example 3, we calculated $s=2.408318916$, and $\sum {\left(x-\overline{x}\right)}^2=330$.

   From the $t$ table, we find that, for 99% confidence, $t_{\alpha /2}$ for $n-2=10-2=8$ degrees of freedom is $t_{\alpha /2}=3.355$. So, our margin of error $E$ si

   $E=t_{\alpha /2}·{\displaystyle \frac{s}{\sqrt{\sum {\left(x-\overline{x}\right)}^2}}}=\left(3.355\right)\left({\displaystyle \frac{2.408318916}{\sqrt{330}}}\right)\approx 0.4448$

   The 99% confidence interval for ${\beta }_1$ is then given by

   $b_1\pm E=-1.5\pm 0.4448=\left(-1.9448,-1.0552\right)$

   We are 99% confident that the interval (−1.9448, −1.0552) captures the slope ${\beta }_1$ of the population regression line. That is, we are 99% confident that, for each additional year of age, the decrease in the number of text messages lies between 1.9448 and 1.0552.

2. The hypotheses are

   $H_0:{\beta }_1=0$ No linear relationship exists between age and text messages.

   $H_a:{\beta }_1\ne 0$ A linear relationship exists between age and text messages.

FIGURE 14 Scatterplot of Scrabble frequency versus English relative frequency of letters, with regression line.

$\hat{y} = 81.5 \left(\text{relative} \text{frequency} \text{in} \text{Engilish}\right) + 0.636$

FIGURE 15 Minitab regression output.

• $H_{0 }: {\beta }_{1 } = 0$ No linear relationship exists between Scrabble frequency and English relative frequency.
• $H_{a }: {\beta }_{1 }\ne 0$ A linear relationship exists between Scrabble frequency and English relative frequency.

FIGURE 16 Scatterplot of Scrabble frequency versus English relative frequency. Linear regression would not be appropriate.

• What is a “good” Scrabble tile to pick up? In the best case, it would be a letter with high English frequency worth lots of Scrabble game points. Unfortunately, the two do not go together. The high-frequency letters such as E and T have low point values, and the high point-value letters such as Q and Z have low frequencies. But we can still make comparisons.
• Which would you rather pick up, a D or a G? It would seem that D would be preferable because it has the same point value as G with much higher English frequency.
• Which do you prefer between J and X? They are worth the same points, but X has a higher frequency in English, so it is easier to make words with it in the game. The letter H seems to have a good combination of high points and moderate frequency.