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1. Perform the $t$ test for the mean using the critical-value method.
2. Perform the $t$ test for the mean using the $p$-value method.
3. Use confidence intervals to perform two-tailed hypothesis tests.

Note: Students may want to review the characteristics of the $t$ distribution on page 450 of Chapter 8.

$t_{\text{data}}=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

The degrees of freedom is a measure of how the $t$ distribution changes as the sample size changes.

• Step 1 State the hypotheses.

  Use one of the forms from Table 8. State the meaning of $\mu$.

• Step 2 Find $t_{\text{crit}}$ and state the rejection rule.

  Use Table D in the Appendix and Table 8.

• Step 3 Calculate $t_{\text{data}}$.

  $t_{\text{data}}=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

• Step 4 State the conclusion and the interpretation.

  If $t_{\text{data}}$ falls within the critical region, then reject $H_0$. Otherwise, do not reject $H_0$. Interpret your conclusion so that a nonspecialist can understand.

Descriptive statistics for anorexia data.

• Step 1 State the hypotheses.

  The key word “decreasing” guides us to state our hypotheses as follows:

  $\begin{array}{lll}{H}_0:\mu =15\hfill & \text{versus}\hfill & H_a\hfill \end{array}:\mu <15$

  where $\mu$ refers to the population mean age at onset.

• Step 2 Find $t_{\text{crit}}$ and state the rejection rule.

  Our hypotheses from Step 1 indicate that we have a left-tailed test, meaning that the critical region represents an area in the left tail (see Figure 20, page 528). To find $t_{\text{crit}}$, we turn to the $t$ table, an excerpt of which is shown in Figure 19. Because we have a one-tailed test, under “Area in one tail,” select the column with our $\alpha$ value 0.05. Then choose the row with our $\text{df}=n-1=20-1=19$, so that we get $t_{\text{crit}}=1.729$. Because we have a left-tailed test, the rejection rule from Table 8 is “Reject $H_0$ if $t_{\text{data}}\le -t_{\text{crit}}$ ”; that is, we will reject $H_0$ if $t_{\text{data}}\le -1.729$.

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  FIGURE 18 Normal probability plot for age at onset of anorexia nervosa.

  
  FIGURE 19 Finding $t_{\text{crit}}$ for a one-tailed test. For a two-tailed test, use “Area in two tails.”

  

• Step 3 Calculate $t_{\text{data}}$.

  We have $n=20$, $\overline{x}=14.250$, and $s=1.512\text{ years}$. Also, ${\mu }_0=15$, because this is the hypothesized value of $\mu$ stated in $H_0$. Therefore, our test statistic is

  ${\displaystyle t_{\text{data}}=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}=\frac{14.250-15}{1.512/\sqrt{20}}\approx -2.2183}$

• Step 4 State the conclusion and interpretation.

  The rejection rule from Step 2 says to reject $H_0$ if $t_{\text{data}}\le -1.729$. From Step 3, we have $t_{\text{data}}=-2.2183$. Because −2.2183 is less than −1.729, our conclusion is to reject $H_0$. If you prefer the graphical approach, consider Figure 20, which shows where $t_{\text{data}}$ falls in relation to the critical region. Because $t_{\text{data}}=-2.2183$ falls within the critical region, our conclusion is to reject $H_0$. There is evidence at level of significance $\alpha =0.05$ that the population mean age of onset has decreased from its previous level of 15 years.

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  FIGURE 20 Our $t_{\text{data}}=-2.2183$ falls in the critical region.

YOUR TURN#8

1. For the data in Example 18, test, using level of significance $\alpha =0.01$, whether the mean age at onset of anorexia nervosa in young women has been decreasing.
2. Discuss two possible resolutions to the contradiction between the conclusions in Example 18 and Part (a).

• Step 1 State the hypotheses.

  The key words “differs from” indicate a two-tailed test, with ${\mu }_0=38$, because we are testing whether $\mu$ differs from 38. So our hypotheses are

  $\begin{array}{lll}{H}_0:\mu =38\hfill & \text{versus}\hfill & H_a:\mu \ne 38\hfill \end{array}$

  where $\mu$ represents the population mean number of hours worked per week.

  

  FIGURE 21 Critical region for two-tailed test.

• Step 2 Find $t_{\text{crit}}$ and state the rejection rule.

  To find $t_{\text{crit}}$ for a two-tailed test with level of significance $\alpha =0.10$, we look in the 0.10 column in the “Area in two tails” section of Table D in the Appendix. The degrees of freedom $\text{df}=n-1=29$ gives us $t_{\text{crit}}=1.699$. From Table 8, the rejection rule is: “Reject $H_0$ if $t_{\text{data}}\ge 1.699$ or $t_{\text{data}}\le -1.699$.”

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• Step 3 Calculate $t_{\text{data}}$:

  $t_{\text{data}}=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}=\frac{35.26-38}{10/\sqrt{30}}\approx -1.5$

• Step 4 State the conclusion and the interpretation.

  $t_{\text{data}}=-1.5$ is not ≥1.699 and it is not ≤−1.699; therefore, we do not reject $H_0$. See Figure 21. There is insufficient evidence, at level of significance $\alpha =0.10$, that the population mean number of hours worked per week differs from 38.

• Step 1 State the hypotheses and the rejection rule.

  Use one of the forms from Table 9. State the meaning of $\mu$. The rejection rule is “Reject $H_0$ if the $p\text{-value}\le \alpha$.”

• Step 2 Calculate $t_{\text{data}}$.

  $t_{\text{data}}=\frac{\overline{\mathrm{x}}-{\mu }_0}{s/\sqrt{n}}$

• Step 3 Find the $p$-value.

  Either use technology to find the $p$-value or estimate the $p$-value using Table D, $t$ Distribution, in the Appendix.

• Step 4 State the conclusion and the interpretation.

  If the $p\text{-value}\le \alpha$, then reject $H_0$. Otherwise, do not reject $H_0$. Interpret your conclusion.

FIGURE 22 Normal probability plot of milk prices.

• Step 1 State the hypotheses and the rejection rule.

  The key words “is greater than” means that we have a right-tailed test. Answering the question “Greater than what?” gives us ${\mu }_0=3.65$.

  $\begin{array}{lll}{H}_0:\mu =3.65\hfill & versus\hfill & H_a:\mu >3.65\hfill \end{array}$

  where $\mu$ represents the population mean price of milk. We will reject $H_0$ if the $p\text{-value}\le \alpha =0.10$.

• Step 2 Calculate $t_{\text{data}}$.

  We use the instructions from the Step-by-Step Technology Guide on page 537. Figure 23 shows the TI-83/84 results from the $t$ test for $\mu$.

  FIGURE 23 TI-83/84 results for right-tailed $t$ test.

  

  For a more accurate calculation of the $p$-value, we retain 9 decimal places for the value of $t_{\text{data}}$.

  Using the statistics from Figure 23, we have the test statistic

  $t_{\text{data}}=\frac{\overline{\mathrm{x}}-{\mu }_0}{s/\sqrt{n}}=\frac{3.714-3.65}{0.0873307888/\sqrt{10}}=2.317461838\approx 2.3175$

• Step 3 Find the $p$-value.

  From Figures 23 and 24, we have

  $\mathrm{p}\text{-value} =P\text{(t }\ge \text{ 2}\text{.317461838) }=\text{ 0}\text{.0228376005}\approx 0.0228$

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  FIGURE 24 The $p$-value for a right-tailed $t$ test.

  

• Step 4 State the conclusion and the interpretation.

  The $\text{p-value}\approx 0.0228$ is less than the level of significance $\alpha =0.05$; therefore, reject $H_0$. There is evidence, at level of significance $\alpha =0.05$, that the population mean price of milk is greater than $3.65.

YOUR TURN#9

1. Hypotheses
2. $t_{\text{data}}$
3. $p$-value
4. Conclusion

FIGURE 25a The golden ratio.

FIGURE 25b The Parthenon follows the golden ratio.

FIGURE 26 Mona Lisa's face follows the golden ratio.

FIGURE 27 Beaded dress of Nahtoma, daughter of Chief Washakie of the Eastern Shoshone, showing rectangles that may follow the golden ratio.

FIGURE 28 Normal probability plot.

• Step 1 State the hypotheses and the rejection rule.

  We are interested in whether the population mean length-to-width ratio of Shoshone beaded rectangles equals the golden ratio of 1.618, so we perform a two-tailed test:

  $H_0:\mu =1.618\hspace{1em}\text{versus}\hspace{0.17em}\hspace{1em}{H}_a:\mu \ne 1.618$

  where $\mu$ represents the population mean length-to-width ratio of Shoshone beaded rectangles. We will reject $H_0$ if the $\text{p-value}\le 0.05$.

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• Step 2 Find $t_{\text{data}}$.

  Using the statistics from Figure 29a, we have the test statistic

  $t_{\text{data}}=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}=\frac{1.583888889-1.618}{0.1230083158/\sqrt{18}}\approx -1.176545482\approx -1.1765$

• Step 3 Find the $p$-value.

  From Figures 29a, 29b, and 30, we have

  $p\text{-value}=P\left(t>|-1.1765|\right)+P\left(t<-|-1.1765|\right)\approx 0.2556$

• Step 4 State the conclusion and interpretation.

  Because $\text{p-value}\approx 0.2556$ is $not\le \alpha =0.05$, we do not reject $H_0$. Thus, there is insufficient evidence, at level of significance $\alpha =0.05$, that the population mean ratio differs from 1.618. In other words, the data do not reject the claim that Shoshone beaded rectangles follow the same golden ratio exhibited by the Parthenon and the Mona Lisa.

  

  FIGURE 29a TI-83/84 results.
  

  FIGURE 29b CrunchIt! results.
  

  FIGURE 30 $p$-Value for two-tailed $t$ test is sum of two tail areas.

$\begin{array}{lll}{H}_0:\mu =38\hfill & \text{versus}\hfill & H_a:\mu \ne 38\hfill \end{array}$

FIGURE 31 Estimating the $p$-value using the $t$ table.

FIGURE 32 Actual $p$-value confirms the estimate.

YOUR TURN#10

$\begin{array}{ll}\text{Lower bound}=144.1\text{ grams}\hfill & \text{Upper bound}=227.7\text{ grams}\hfill \end{array}$

1. $H_0:\mu =100\text{versus}{H}_a:\mu \ne 100$

   The confidence interval is (144.1, 227.7), and because ${\mu }_0=100$ lies outside the interval (see Figure 33), we reject $H_0$.

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2. $\begin{array}{lll}{H}_0:\mu =170\hfill & \text{versus}\hfill & H_a:\mu \ne 170\hfill \end{array}$

   ${\mu }_0=170$ lies inside the interval, so we do not reject $H_0$.

3. $\begin{array}{lll}{H}_0:\mu =250\hfill & \text{versus}\hfill & H_a:\mu \ne 250\hfill \end{array}$

   ${\mu }_0=250$ lies outside the interval, so we reject $H_0$.

   

   FIGURE 33 Reject $H_0$ for values of ${\mu }_0$ that lie outside (144.1, 227.7).

• Step 1 State the hypotheses and the rejection rule.
• Step 2 Find $t_{\text{data}}$.
• Step 3 Find the $p$-value.
• Step 4 State the conclusion and the interpretation.

1. SPSS output for a $t$ test for $\mu$, where $\mu$ represents the population mean number of orchard farms per county, nationwide.

   

2. JMP output for a $t$ test for $\mu$, where $\mu$ represents the population mean number of grocery stores per county, nationwide.

   

1. Interpreting the SPSS output.

   • Step 1 State the hypotheses and the rejection rule.

     In the SPSS output, the “Test $\text{Value}=40$” indicates that ${\mu }_0=40$. Also, the “2-tailed” in the output indicates that we have a two-tailed test. Thus, our hypotheses are:

     $\begin{array}{lll}{H}_0:\mu =40\hfill & \text{versus}\hfill & H_a:\mu \ne 40\hfill \end{array}$

     where $\mu$ represents the population mean number of orchard farms per county, nationwide. We will reject $H_0$ if the $p$-value is less than level of significance $\alpha =0.10$.

   • Step 2 Find $t_{\text{data}}$.

     Under the “t” in the SPSS is the value for $t_{\text{data}},-1.079$.

   • Step 3 Find the $p$-value.

     The abbreviation “Sig.” stands for “Significance,” which represents the $p$-value: 0.281.

   • Step 4 State the conclusion and the interpretation.

     The $p$-value of 0.281 is not less than the level of significance $\alpha =0.10$, so we do not reject $H_0$. There is insufficient evidence that the population mean number of orchard farms per county differs from 40.

2. Interpreting the JMP output.

   • Step 1 State the hypotheses and the rejection rule.

     In the JMP output, the “Hypothesized Value” indicates that ${\mu }_0=20$. Now, JMP is unusual in that it performs all three types of hypothesis test simultaneously: two-tailed, right-tailed, and left-tailed, as shown in the JMP output. It does not specify a particular form of the test. Let us use the right-tailed test for this example. Thus, our hypotheses are:

     $\begin{array}{lll}{H}_0:\mu =20\hfill & \text{vs}\hfill & H_a:\mu >20\hfill \end{array}$

     where $\mu$ represents the population mean number of grocery stores per county, nationwide. We will reject $H_0$ if the $p$-value is less than level of significance $\alpha =0.10$.

   • Step 2 Find $t_{\text{data}}$.

     Next to “Test Statistic” in the JMP output, we find the value of our test statistic, $t_{\text{data}}$, 0.3324.

   • Step 3 Find the $p$-value.

     Here, we need to be careful, because JMP gives us three different $p$-values, depending on which form of the hypothesis test is performed. We chose the right-tailed test, so our $p$-value is next to $“\text{Prob}>t”:p\text{-value}=0.3698$, as indicated in the JMP output.

   • Step 4 State the conclusion and the interpretation.

     The $p$-value of 0.3698 is not less than the level of significance $\alpha =0.10$, so we do not reject $H_0$. There is insufficient evidence that the population mean number of grocery stores per county is greater than 20.