1. Perform the $Z$ test for the mean, using the $p$-value method.
2. Assess the strength of evidence against the null hypothesis.
3. Describe the relationship between the $p$-value method and the critical-value method.
4. Use the $Z$ confidence interval for the mean to perform the two-tailed $Z$ test for the mean.

Remember that probability is represented by the area under the curve.

1. $H_0:\mu =3.0\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu >3.0,Z_{\text{data}}=1$
2. $H_0:\mu =10\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <10,Z_{\text{data}}=-1.45$
3. $H_0:\mu =100\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 100,Z_{\text{data}}=-2$

1. We have a right-tailed test, so that the $p$-value equals the area in the right tail:

   $p\text{-value}=P\left(Z>Z_{\text{data}}\right)=P\left(Z>1\right)$

   The $Z$ table gives the probability for $P\left(Z<1\right)$. Thus,

   $p\text{-value}=P\left(Z>1\right)=1-P\left(Z<1\right)=1-0.8413=0.1587$ (Figure 6a)

2. We have a left-tailed test, so that the $p$-value equals the area in the left tail:

   $p\text{-value}=P\left(Z<Z_{\text{data}}\right)=P\left(Z<-1.45\right)=0.0735$ (Figure 6b)

   Page 509

3. Here, we have a two-tailed test, so that the $p$-value equals the sum of the areas in the two tails:

   $\begin{array}{c}p\text{-value}=P\left(Z>\left|Z_{\text{data}}\right|\right)+\left(Z<-\left|Z_{\text{data}}\right|\right)\\ =P\left(Z>\left|-\hspace{0.17em}2\right|\right)+\left(Z<-\left|-\hspace{0.17em}2\right|\right)\\ =P\left(Z>2\right)+\left(Z<-2\right)\\ =0.0228+0.0228=0.0456\hspace{0.17em}\left(\text{Figure}\hspace{0.17em}6\mathrm{c}\right)\end{array}$

   

   FIGURE 6a $p$-Value for a right-tailed test.
   

   FIGURE 6b $p$-Value for a left-tailed test.
   

   FIGURE 6c $p$-Value for a two-tailed test.

YOUR TURN#5

1. $H_0:\mu =75\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu >75,Z_{\text{data}}=0.5$
2. $H_0:\mu =50\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <50,Z_{\text{data}}=-1.2$
3. $H_0:\mu =1\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 1,Z_{\text{data}}=-0.1$

The $p$-Value applet allows you to experiment with various hypotheses, means, standard deviations, and sample sizes in order to see how changes in these values affect the $p$-value.

This rejection rule can be applied to any type of hypothesis test we perform in Chapters 9–14 using the $p$-value method.

• Step 1 State the hypotheses and the rejection rule.

  Use one of the forms from Table 5 to write the hypotheses. State the meaning of $\mu$. The rejection rule is “Reject $H_0$ if the $p$-value ≤ $\alpha$.”

• Step 2 Calculate $Z_{\text{data}}$.

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

  where the sample mean $\overline{x}$ and the sample size $n$ represent the sample data, and the population standard deviation σ represents the population data.

• Step 3 Find the $p$-value.

  Either use technology to find the $p$-value, or calculate it using the form in Table 5 that corresponds to your hypotheses.

• Step 4 State the conclusion and interpretation.

  If the $p$-value ≤ $\alpha$, then reject $H_0$. Otherwise do not reject $H_0$. Interpret your conclusion so that a nonspecialist (someone who has not had a course in statistics) can understand, as follows:

• Interpretation when you reject $H_0$: There is evidence at level of significance $\alpha$ that [whatever ${\mathrm{H}}_{\mathrm{a}}$ says].
• Interpretation when you do not reject $H_0$: There is insufficient evidence at level of significance $\alpha$ that [whatever ${\mathrm{H}}_{\mathrm{a}}$ says].

• Step 1 State the hypotheses and the rejection rule.

  The key words here are “has fallen,” which means “is less than.” The answer to the question “Less than what?” gives us ${\mu }_0=3.0$. Thus, our hypotheses are

  $H_0:\mu =3.0\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <3.0$

  where $\mu$ refers to the population mean user rating for JFK Airport. We will reject $H_0$ if the $p\text{-value}\le \alpha =0.05$.

• Step 2 Calculate $Z_{\text{data}}$.

  We have $\overline{x}=2.75,{\mu }_0=3.0,n=36,$, and $\sigma =1$. Thus, our test statistic is

  $Z_{\text{data}}=\frac{\overline{x}-{\mu }_0}{\sigma /\sqrt{n}}=\frac{2.75-3.0}{1/\sqrt{36}}=-1.5$

  Page 511

• Step 3 Find the $p$-value.

  Our hypotheses represent a left-tailed test from Table 5. Thus,

  $p\text{-value}=P\left(Z<Z_{\text{data}}\right)=P\left(Z<-1.5\right)$

  This is a Case 1 problem from Table 8 in Chapter 6 (page 355). The $Z$ table (Appendix Table C) provides us with the area to the left of $Z=-1.5$ (Figure 7):

  $P\left(Z<-1.5\right)=0.0668$

  Thus, the $p$-value is 0.0668.

  

  FIGURE 7 The $p$-value 0.0668 is not ≤ 0.05, so do not reject $H_0$.

• Step 4 State the conclusion and interpretation.

  Our level of significance is $\alpha =0.05$ (from Step 1). The $p\text{-value}=0.0668$ is not ≤ 0.05, therefore, we do not reject $H_0$. There is insufficient evidence at the level of significance $\alpha =0.05$ that the population mean user rating for JFK Airport is less than 3.0.

Note from Figure 8 that the sample mean birth weight $\overline{x}=3276$ grams is close to the hypothesized mean birth weight of ${\mu }_0=3200$ grams. This value of $\overline{x}$ is not extreme and thus does not seem to offer strong evidence that the hypothesized mean birth weight is wrong. Therefore, we might expect to not reject the hypothesis that ${\mu }_0=3200$ grams.

FIGURE 8 Sample mean, $\overline{x}=3276$, is close to hypothesized mean, ${\mu }_0=3200$, so we expect to not reject the null hypothesis.

• Step 1 State the hypotheses and the rejection rule.

  The key words “differs from” mean that we have a two-tailed test:

  $H_0:\mu =3200\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 3200$

  where $\mu$ refers to the population mean birth weight of Brisbane babies. We will reject $H_0$ if the $p\text{-value}\le \alpha =0.10$.

• Step 2 Calculate $Z_{\text{data}}$.

  We will use the instructions provided in the Step-by-Step Technology Guide at the end of this section (page 519). Figure 9 shows the TI-83/84 results from the $Z$ test for $\mu$:

  FIGURE 9 TI-83/84 results.

  
  Page 513

  $Z_{\text{data}}=\frac{\overline{x}-{\mu }_0}{\sigma /\sqrt{n}}=\frac{3276-3200}{528/\sqrt{44}}=0.9547859245\approx 0.9548$

• “Test of $\mu =3200$ versus ≠ 3200” refers to the hypotheses being tested, $H_0:\mu =3200$ versus $H_a:\mu \ne 3200$.
• “The assumed standard deviation = 528” refers to our assumption that $\sigma =528$.
• SE Mean refers to the standard error of the mean, that is, $\sigma /\sqrt{n}$. You can see that $528/\sqrt{44}\approx 79.6$.
• 90% CI represents a 90% $Z$ confidence interval for $\mu$.

• $Z$ refers to our test statistic:

  $Z_{\text{data}}=\frac{\overline{x}-{\mu }_0}{\sigma /\sqrt{n}}=\left(3276-3200\right)/\left(528/\sqrt{44}\right)=0.9547859245\approx 0.95$

• $P$ represents our $p$-value of 0.340.
  FIGURE 10 Minitab results.

  

FIGURE 11 JMP results.

• “Hypothesized Value” refers to the hypotheses being tested: $H_0:\mu =3200$ versus $H_a:\mu \ne 3200$.
• “Actual Estimate” refers to the sample mean, $\overline{x}=3276$.
• “Sigma given” refers to our assumption that $\sigma =528$.
• “Test statistic” refers to $Z_{\text{data}}$, our test statistic.
• “$\text{Prob}>\left|\mathrm{z}\right|$,” “$\text{Prob}>\mathrm{z}$,” and “$\text{Prob}<\mathrm{z}$” refers to the $p$-value of a two-sided, right-tailed, and left-tailed test, respectively. We want the two-sided $p$-value, which is $“\text{Prob}>\left|\mathrm{z}\right|”=0.3397$.

• Step 3 Find the $p$-value.

  We have a two-tailed test from Step 1, so that from Table 5 our $p$-value is (Figure 12)

  $\begin{array}{c}p\text{-Value}=2·P\left(Z>\left|{\mathrm{Z}}_{\text{data}}\right|\right)=2·P\left(Z>0.9548\right)\approx 2·\left(0.1698\right)\\ =0.3396\end{array}$

  FIGURE 12 $p$-Value is the sum of two tail areas: $0.1698+0.1698=0.3396$.

  
  Page 514

• Step 4 State the conclusion and interpretation.

  Because 0.3396 is not ≤ 0.10, we do not reject $H_0$. There is insufficient evidence that the population mean birth weight differs from 3200 grams. This conclusion is just as we expected.

The $p$-value provides us with the smallest level of significance at which the null hypothesis would be rejected, that is, the smallest value of $\alpha$ at which the results would be considered significant.

Note: Use Table 6 for all exercises that ask for an assessment of the strength of evidence against the null hypothesis.

1. In Example 13, we tested $H_0:\mu =3.0$ versus $H_a:\mu <3.0$, where $\mu$ refers to the population mean user rating for JFK International Airport. Our $p$-value of 0.0668 implies that there is moderate evidence against the null hypothesis that the population mean user rating for JFK Airport equals 3.0.

   Page 515
2. In Example 14, we tested $H_0:\mu =3200$ versus $H_a:\mu \ne 3200$, where $\mu$ refers to the population mean birth weight of Brisbane babies (in grams). Our $p$-value of 0.3397 implies that there is no evidence against the null hypothesis that the population mean birth weight of Brisbane babies equals 3200 grams.

YOUR TURN#6

1. $H_0:\mu =3.0\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu >3.0,p\text{-value}=0.1587$
2. $H_0:\mu =10\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <10,p\text{-value}=0.0735$
3. $H_0:\mu =100\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 100,p\text{-value}=0.0456$

Developing Your Statistical Sense

1. We don't want the choice of a to dictate our conclusion, so perhaps we should turn to a direct assessment of the strength of evidence against the null hypothesis, as provided in Table 6. In this case, the $p$-value of about 0.0668 would offer moderate evidence against the null hypothesis, regardless of the value of $\alpha$.
2. Obtain more data, perhaps through a call for further research.

• If a certain value for ${\mu }_0$ lies outside the corresponding $100\left(1-\alpha \right)\%$$Z$ confidence interval for $\mu$, then the null hypothesis specifying this value for ${\mu }_0$ would be rejected for level of significance $\alpha$ (see Figure 15).
• Alternatively, if a certain value for ${\mu }_0$ lies inside the $100\left(1-\alpha \right)\%$$Z$ confidence interval for $\mu$, then the null hypothesis specifying this value for ${\mu }_0$ would not be rejected for level of significance $\alpha$.

• If any values of ${\mu }_0$ lie inside the confidence interval, that is, between 471.2 and 548.8, we will not reject $H_0$ for this value of ${\mu }_0$.
• If any values of ${\mu }_0$ lie outside the confidence interval, that is, either to the left of 471.2 or to the right of 548.8, we will reject $H_0$, as shown in Figure 16.
  FIGURE 16 Reject $H_0$ for values of ${\mu }_0$ that lie outside (471.2, 548.8).

  

1. $H_0:\mu =470\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 470$
2. $H_0:\mu =510\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 510$
3. $H_0:\mu =550\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 550$

YOUR TURN#7

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

1. TI-83/84 output for a Z test for $\mu$, where $\mu$ represents the population mean length of laboratory mice (in cm)
2. Minitab output for a $Z$ test for $\mu$, where $\mu$ represents the population mean number of farmer's markets per county, nationwide.
   

   TI-83/84 output for part (a).
   

   Minitab output for part (b).

1. Interpreting the TI-83/84 output.

   • Step 1 State the hypotheses and the rejection rule.

     In the TI-83/84 output, the “$\mu >10$” indicates the alternative hypothesis. In other words, the hypotheses are:

     $H_0:\mu =10\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu >10$

     where $\mu$ represents the population mean length of laboratory mice (in cm). We will reject $H_0$ if the $p$-value is less than the level of significance $\alpha =0.05$.

     Page 519

   • Step 2 Find $Z_{\text{data}}$.

     The “$\mathrm{z}=1$” in the TI-83/84 output provides us the value of the test statistic, $Z_{\text{data}}=1$.

   • Step 3 Find the $p$-value.

     The “$p=.1586552596$” in the TI-83/84 output represents the $p$-value.

   • Step 4 State the conclusion and interpretation.

     The $p$-value from Step 3 is not less than the level of significance $\alpha =0.05$, so we do not reject $H_0$. There is insufficient evidence that the population mean length of laboratory mice is greater than 10 cm.

2. Interpreting the Minitab output.

   • Step 1 State the hypotheses and the rejection rule.

     The first line in the Minitab output is “Test of $\mu =2.3\hspace{0.17em}vs\ne 2.3$,” which indicates a two-tailed test, as follows:

     $H_0:\mu =2.3\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu \ne 2.3$

     where $\mu$ represents the population mean number of farmer's markets per county. We will reject $H_0$ if the $p$-value is less than the level of significance $\alpha =0.05$.

   • Step 2 Find $Z_{\text{data}}$

     Under “$Z$” in the Minitab output is found “2.91,” giving us $Z_{\text{data}}=2.91$.

   • Step 3 Find the $p$-value.

     Under “$p$” in the Minitab output is “0.004,” representing our $p$-value.

   • Step 4 State the conclusion and interpretation.

     The $p$-value (0.004) from Step 3 is less than the level of significance $\alpha =0.05$, so we reject $H_0$.There is evidence that the population mean number of farmer's markets per county, nationwide, differs from 2.3.