Problem

Claim: The population mean is at least 9.

Sample Properties: \(n\) = 21, \(\bar{x}\) = 9.35, \(s_{X}\) = 3.46

Steps

1. Claim: “The population mean is at least 9.”

“At least 9” means “9 or more,” so this translates to the claim: \[ \mu_{X} \geq 9.\]

2. The claim includes equality, so is one of \(\leq, \geq, =\), and thus the null hypothesis.

3. The opposite of “at least 9” is “less than 9,” so the complementary hypothesis is \[ \mu_{X} < 9.\]

4. The null and alternative hypotheses are: \[ \begin{aligned} H_{0} &: \mu_{X} \geq 9 \\ H_{a} &: \mu_{X} < 9. \end{aligned}\]

5. Values of \(\bar{x}\) less than 9 would be evidence against the null hypothesis.

6. Under the null hypothesis that \(\mu_{X} = 9,\) \[ T = \frac{\bar{X} - 9}{S_{X}/\sqrt{n}} \sim t(n - 1),\] which has the density curve

7. \[ t_{\text{obs}} = \frac{\bar{x} - 9}{s_{X}/\sqrt{n}} = \frac{9.35 - 9}{3.46/\sqrt{21}} = 0.4635553\]

8. A negative \(T\)-statistic would be evidence against the null, and when the null hypothesis is true, we get a \(T\)-statistic this negative or more negative with probability \[ P(T \leq t_{\text{obs}}) = P(T \leq 0.4635553) = \texttt{pt(0.4635553, 21 - 1)} = 0.6760142.\] so this is not very strong evidence against the null hypothesis.