1. Describe the characteristics of the $F$ distribution and the $F$ test for two population standard deviations.
2. Perform hypothesis tests for two population standard deviations using the critical-value method.
3. Perform hypothesis tests for two population standard deviations using the $p$-value method.

$F_{\text{data}}=\frac{s_1^2}{s_2^2}$

1. The total area under the $F$ curve equals 1.
2. The value of the $F$ random variable is never negative, so the $F$ curve starts at 0. However, it extends indefinitely to the right. The curve approaches but never quite meets the horizontal axis.
3. Because of the characteristics described in (2), the $F$ curve is right-skewed.
4. There is a different $F$ curve for each different pair of degrees of freedom: ${\text{df}}_1$ and ${\text{df}}_2$.

• Step 1 Look across the top of the $F$ table until you find your ${\text{df}}_1$. Then go down that column until you see your ${\text{df}}_2$ on the left.
• Step 2 For each ${\text{df}}_2$ on the left, you will see a range of $\alpha$ values from 0.100 to 0.001. Choose the row next to ${\text{df}}_2$ that has your value of $\alpha$. The $F$-value in that row and column is your value of $F{}_{\text{crit}}$.

Note: When the degrees of freedom are not listed in the $F$ table, we do not necessarily take the closest degrees of freedom we can find. This is because, sometimes, the closest degree of freedom is larger than the original, which leads to misleadingly overprecise results—α level of precision not warranted by the data. For example, this could lead us to find significance where none actually exists.

Developing Your Statistical Sense

1. Switch the values of ${\text{df}}_1$ and ${\text{df}}_2$.
2. Find $F_{\alpha ,n_2-1,n_1-1}$ using the $F$ table.
3. Calculate $F_{\text{crit}}=F_{1-\alpha ,n_1-1,n_2-1}=\frac{1}{F_{\alpha ,n_2-1,n_1-1}}$

FIGURE 21 $F$ table (excerpt).

1. Find the critical value with area $\alpha =0.05$ to the right of it, for an $F$ distribution with ${\text{df}}_1=2$ and ${\text{df}}_2=7$.
2. Find the critical value with area $\alpha =0.01$ to the left of it, for an $F$ distribution with ${\text{df}}_1=6$ and ${\text{df}}_2=3$.

1. • Step 1. Go across the top of the $F$ table until we get to ${\text{df}}_1=2$, and go down that column until we see ${\text{df}}_2=7$ on the left.

   • Step 2. Next to the 7 is a range of $\alpha$ values from 0.100 to 0.001. We choose the row with $\alpha =0.05$. The $F$-value in that row and column is 4.74. Thus, our $F$ critical value is

     $F_{\text{crit}}=F_{0.05,2,7}=4.74$

     Page 621
2. • Step 1. Switching the two degrees of freedom, we go across the top of the $F$ table until we get to ${\text{df}}_1=3$. Then we go down that column until we see ${\text{df}}_2=6$ on the left.
   • Step 2. Choose the row with $\alpha =0.010=0.01$. The $F$-value in that row and column is 9.78. Thus, our $F$ critical value is
     

     $F_{\text{crit}}=F_{0.99,6,3}=\frac{1}{F_{0.01,6,3}}=\frac{1}{9.78}=0.1023$

• Step 1 State the hypotheses. Use one of the forms from Table 17. State the meaning of ${\sigma }_1$ and ${\sigma }_2$.
• Step 2 Find the critical value(s) and state the rejection rule. Use Table 17 and the $F$ tables.

• Step 3 Find $\mathit{F}{}_{\mathbf{\text{data}}}$.

  $F_{\text{data}}=\frac{s_1^2}{s_2^2}$

  follows an $F$ distribution with ${\text{df}}_1=n_1-1$ and ${\text{df}}_2=n_2-1$.

• Step 4 State the conclusion and the interpretation. Compare $F_{\text{data}}$ with the $F$ critical value from Table 17.

FIGURE 22a Normal probability plot of Google stock prices.

FIGURE 22b Normal probability plot of Apple stock prices.

• Step 1 State the hypotheses. We are testing whether Google's stock prices are more variable than those of Apple. Thus, because Google represents population 1, we have the following hypotheses for our $F$ test:

  $\begin{array}{ccc}{H}_0:{\sigma }_1={\sigma }_2& \text{versus}& H_a:{\sigma }_1>{\sigma }_2\end{array}$

  where ${\sigma }_1$ represents the standard deviation of Google stock prices and ${\sigma }_2$ represents the standard deviation of Apple stock prices. Use level of significance $\alpha =0.05$.

• Step 2 Find the critical value and state the rejection rule. We have ${\text{df}}_1=n{}_1-1=7$ and ${\text{df}}_2=n_2-1=6$. From Table 17 and Appendix Table F, our critical value is the $F$-value with area $\alpha =0.05$ to the right of it:

  $F_{\text{crit}}=F_{\alpha ,n_1-1,n_2-1}=F_{0,0.5,7.6}=4.21$

  Our rejection rule is, therefore, from Table 17: Reject $H_0$ if $F_{\text{data}}>4.21$.

  Page 623

• Step 3 Find ${\mathit{F}}_{\mathbf{\text{data}}}$

  $F_{\text{data}}=\frac{S_1^2}{S_2^2}=\frac{{7.999701}^2}{{1.862594}^2}\approx 18.45$

  follows an $F$ distribution with ${\text{df}}_1=n_1-1=n_1-1=7$ and ${\text{df}}_2=n_2-1=n_2-1=6$.

• Step 4 State the conclusion and the interpretation. Because $F_{\text{data}}\approx 18.45$ is greater than $F_{\text{crit}}=4.21$, we reject $H_0$. There is evidence that the variability in Google stock prices is greater than the variability in Apple stock prices.

• Step 1 State the hypotheses and the rejection rule. Use one of the forms from Table 19. Clearly state the meaning of ${\sigma }_1$ and ${\sigma }_2$. The rejection rule is: Reject ${\mathrm{H}}_0$ if the $\mathrm{p}$-value is less than $\alpha$.

• Step 2 Find ${\mathit{F}}_{\mathbf{\text{data}}}$.

  $F_{\text{data}}=\frac{S_1^2}{S_2^2}$

  follows an $F$ distribution with ${\text{df}}_1=n{}_1-1$ and ${\text{df}}_2=n{}_2-1$.

• Step 3 Find the $\mathbf{p}$ -value. Use technology and Table 19 to find the $p$-value.
• Step 4 State the conclusion and the interpretation. Compare the $p$-value with $\alpha$.

FIGURE 23a Normal probability plot of Florida percentages.

FIGURE 23b Normal probability plot of Georgia percentages.

• Step 1 State the hypotheses. We are testing whether there is a difference in the standard deviation of the percentages for Florida (${\sigma }_1$) and Georgia (${\sigma }_2$). We therefore have a two-tailed test:

  $\begin{array}{ccc}{H}_0:{\sigma }_1={\sigma }_2& \text{versus}& H_a:{\sigma }_1\ne {\sigma }_2\end{array}$

  where ${\sigma }_1$ and ${\sigma }_2$ represent the standard deviations of the percent of Florida and Georgia respondents, respectively, who reported that their nurses always communicated well. Use level of significance $\alpha =0.10$.

• Step 2 Find ${\mathit{F}}_{\mathbf{\text{data}}}$.

  $F_{\text{data}}=\frac{s_1^2}{s_2^2}=\frac{{3.1305}^2}{{3.8162}^2}\approx 0.6729$

  follows an $F$ distribution with ${\text{df}}_1=n_1-1=10$ and ${\text{df}}_2=n_2-1=10$.

• Step 3 Find the $p$-value. Because we have a two-tailed test, Table 19 states that the $\mathit{p}$-value is

  $\begin{array}{ccc}\text{the smaller of }\left(i\right)2\cdot P(F>F_{\text{data}})& \text{and}& \left(ii\right)2\cdot P(F<F_{\text{data}})\end{array}$

  Figure 24a shows the output from the TI-83/84, giving $P(F>F_{\text{data}})$ as the area under the $F$ distribution curve between 0.6729 and infinity. This gives

  $2\cdot P(F>F_{\text{data}})=2\cdot 0.7287420993=1.457484199$

  

  FIGURE 24a

  This cannot represent a valid $p$-value, because it is larger than 1. Figure 24b shows the output from the TI-83/84, giving $P(F<F_{\text{data}})$ as the area under the $F$ distribution curve between 0 and 0.6729. Thus, our $p$-value is:

  $p\text{-value}=2\cdot P(F<F_{\text{data}})=2\cdot P(F<0.6729)=2\cdot 0.2712579007\approx 0.5425$

  because this is smaller than 1.457484199.

  

  FIGURE 24b
  Page 625
• Step 4 State the conclusion and the interpretation. The $p$-value 0.5425 is not less than $\alpha =0.10$, so we do not reject $H_0$. There is insufficient evidence for a difference in population standard deviations between percentages of patients in Florida and Georgia hospitals who reported that their nurses always communicated well.