November 19

Null and Alternative Hypotheses

In any formal hypothesis test we set up two competing statements about a population parameter:

• The null hypothesis, $H_0$, is the statement that represents the status quo or no effect. It always includes an equality (e.g., $H_0: \mu = \mu_0$, or $H_0: p = p_0$).
• The alternative hypothesis, $H_a$, is the statement we seek evidence for. It does not contain equality and may be two-sided or one-sided:
  • Two-sided: $H_a: \theta \neq \theta_0$.
  • One-sided (right): $H_a: \theta > \theta_0$.
  • One-sided (left): $H_a: \theta < \theta_0$.

Choice of $H_a$ should reflect the scientific question or the direction of interest before seeing the data.

Significance Level ($\alpha$)

The significance level $\alpha$ is the threshold for declaring a result “statistically significant.” It is the long-run probability of making a Type I error when $H_0$ is true. Common choices are $\alpha=0.05,\;0.01,$ or $0.10$. Formally:

\[\alpha = P(\text{Reject } H_0 \mid H_0 \text{ is true}).\]

The critical value(s) for the test statistic are chosen so that the probability of falling into the rejection region under $H_0$ equals $\alpha$ (split between tails for two-sided tests).

Types of Errors

There are two possible incorrect decisions in hypothesis testing.

• Type I error: Rejecting $H_0$ when $H_0$ is true.

  \[P(\text{Type I error}) = \alpha.\]

• Type II error: Failing to reject $H_0$ when $H_a$ is true (i.e., there is a real effect).

  \[\beta = P(\text{Fail to reject } H_0 \mid H_a \text{ is true}).\]

Power of a test is the complement of the Type II error rate:

\[ext{Power} = 1-\beta = P(\text{Reject } H_0 \mid H_a \text{ is true}).\]

Trade-offs and Practical Notes

• Decreasing $\alpha$ (making the test more conservative) reduces the probability of a Type I error but generally increases $\beta$ (reduces power) for a fixed sample size.
• Increasing sample size reduces the standard error, which typically decreases $\beta$ and increases power for fixed $\alpha$.
• Effect size, variability, sample size, and $\alpha$ together determine power. For planning, one often specifies desired power (e.g., 0.80), an effect size of interest, and $\alpha$ to solve for required sample size.

Decision and p-value

Compute a test statistic (e.g., $t$ or $z$) and its p-value. The p-value is defined as:

\[ext{p-value} = P(\text{Data as extreme or more extreme} \mid H_0 \text{ true}).\]

• If $\text{p-value} \le \alpha$, reject $H_0$.
• If $\text{p-value} > \alpha$, fail to reject $H_0$.

Always interpret results in context and report $\alpha$, test statistic, degrees of freedom (if applicable), p-value, and a conclusion about the hypotheses.

Examples

Example 1: Testing a Population Mean

A researcher wants to test if the average weight of a certain species of fish is different from 5 kg. A random sample of 30 fish is taken, and the sample mean weight is 5.3 kg with a standard deviation of 0.8 kg. Test at the 0.05 significance level.

Step 1: State the hypotheses

• Null hypothesis: $H_0: \mu = 5$
• Alternative hypothesis: $H_a: \mu \neq 5$

Step 2: Choose the significance level

• $\alpha = 0.05$

Step 3: Compute the test statistic

• Test statistic: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{5.3 - 5}{0.8 / \sqrt{30}} \approx 2.19$

Step 4: Find the critical value or p-value

• Degrees of freedom: $n - 1 = 29$
• Two-tailed test: Critical $t$-value for $\alpha = 0.05$ is approximately $\pm 2.045$.
• Since $t = 2.19 > 2.045$, reject $H_0$.

Conclusion: There is evidence at the 0.05 significance level to conclude that the average weight of the fish is different from 5 kg.

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Example 2: Testing a Proportion

A company claims that 80% of its customers are satisfied with their service. A random sample of 100 customers shows that 72 are satisfied. Test the claim at the 0.01 significance level.

Step 1: State the hypotheses

• Null hypothesis: $H_0: p = 0.8$
• Alternative hypothesis: $H_a: p \neq 0.8$

Step 2: Choose the significance level

• $\alpha = 0.01$

Step 3: Compute the test statistic

• Test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.72 - 0.8}{\sqrt{0.8(1-0.8)/100}} \approx -2.24$

Step 4: Find the critical value or p-value

• Two-tailed test: Critical $z$-values for $\alpha = 0.01$ are approximately $\pm 2.576$.
• Since $z = -2.24$ is within $[-2.576, 2.576]$, fail to reject $H_0$.

Conclusion: There is not enough evidence at the 0.01 significance level to conclude that the proportion of satisfied customers is different from 80%.

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Example 3: Power of a Test

A clinical trial is designed to detect a mean difference of 5 units in blood pressure with a standard deviation of 10 units. The sample size is 50, and the significance level is 0.05. What is the power of the test if the true mean difference is 5 units?

Step 1: Compute the test statistic under $H_0$

• $t = \frac{\mu - \mu_0}{s / \sqrt{n}} = \frac{5 - 0}{10 / \sqrt{50}} \approx 3.54$

Step 2: Find the critical value

• One-tailed test: Critical $t$-value for $\alpha = 0.05$ and $df = 49$ is approximately 1.676.

Step 3: Compute the power

• Power is the probability of rejecting $H_0$ when $H_a$ is true.
• Using statistical software or tables, the power is approximately 0.95.

Conclusion: The test has a high power of 0.95 to detect the specified mean difference.