To address the hypothesis testing problem, we proceed step-by-step. Here's the solution:

Step 1: State the Problem

We are testing the null hypothesis $H_0: \mu \geq 20$ against the alternative hypothesis $H_1: \mu < 20$. This is a one-tailed test with a significance level of $\alpha = 0.01$.

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Step 2: a. State the Decision Rule

1. The test statistic is a $t$-statistic: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ where $\bar{x}$ is the sample mean, $\mu_0 = 20$ is the hypothesized mean, $s$ is the sample standard deviation, and $n = 5$ is the sample size.

2. The critical value for $t$ is found from the $t$-distribution with $n - 1 = 4$ degrees of freedom at $\alpha = 0.01$. For a one-tailed test, the critical value is $t_{0.01, 4} \approx -3.747$.

Decision Rule: Reject $H_0$ if $t < -3.747$.

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Step 3: b. Compute the Test Statistic

1. Compute the sample mean ($\bar{x}$): $\bar{x} = \frac{18 + 15 + 12 + 19 + 21}{5} = \frac{85}{5} = 17.$

2. Compute the sample standard deviation ($s$): $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}.$ First, find deviations: $x_i - \bar{x} = 18 - 17, \, 15 - 17, \, 12 - 17, \, 19 - 17, \, 21 - 17 \implies 1, \, -2, \, -5, \, 2, \, 4.$ Square deviations: $(x_i - \bar{x})^2 = 1^2, \, (-2)^2, \, (-5)^2, \, 2^2, \, 4^2 \implies 1, \, 4, \, 25, \, 4, \, 16.$ Sum squared deviations: $\sum (x_i - \bar{x})^2 = 1 + 4 + 25 + 4 + 16 = 50.$ Compute $s$: $s = \sqrt{\frac{50}{4}} = \sqrt{12.5} \approx 3.536.$

3. Compute the $t$-statistic: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{17 - 20}{3.536 / \sqrt{5}} = \frac{-3}{3.536 / 2.236} = \frac{-3}{1.581} \approx -1.90.$

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Step 4: c. Decision Regarding $H_0$

The calculated $t$-statistic is $t = -1.90$, which is not less than the critical value $-3.747$. Therefore, we fail to reject the null hypothesis $H_0$.

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Step 5: d. Compute the p-Value

The $t$-statistic of $-1.90$ corresponds to a one-tailed p-value. Using a $t$-distribution table or software, the p-value for $t = -1.90$ with 4 degrees of freedom is approximately: $p \approx 0.070.$ Since $p = 0.070 > 0.01$, we fail to reject $H_0$.

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Conclusion

• a. Decision Rule: Reject $H_0$ if $t < -3.747$.
• b. Test Statistic: $t \approx -1.90$.
• c. Decision: Fail to reject $H_0$. There is insufficient evidence to conclude that the population mean is less than 20 at the $\alpha = 0.01$ level.
• d. p-Value: $p \approx 0.070$.

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Follow-Up Questions:

1. How does the choice of significance level $\alpha$ affect the conclusion in hypothesis testing?
2. What assumptions must hold true for this $t$-test to be valid?
3. How would the result change if the sample size were larger?
4. Can you explain the difference between a one-tailed and a two-tailed test?
5. How does the variability of the data influence the test statistic?

Tip: When working with small sample sizes, always ensure that the population is approximately normal to justify using the $t$-distribution.