Hypothesis Testing for a Single Normal Population

Testing the Mean of a Single Normal Population

When the Variance Is Known (Z Test)

Population: $X \sim N(\mu, \sigma^2)$, with $\sigma^2$ known

Hypotheses:

• $H_0: \mu = \mu_0$
• $H_1: \mu \neq \mu_0$ (two-sided test)
• $H_1: \mu > \mu_0$ (right one-sided test)
• $H_1: \mu < \mu_0$ (left one-sided test)

Test statistic: $Z = \frac{\overline{X} - \mu_0}{\sigma/\sqrt{n}} \sim N(0,1)$

Rejection region:

• Two-sided test: $|Z| > z_{\alpha/2}$
• Right one-sided test: $Z > z_{\alpha}$
• Left one-sided test: $Z < -z_{\alpha}$

Example 1: the lengths of the parts produced by a factory follow the normal distribution $N(\mu, 0.04)$. 16 parts are drawn at random from the products, and the measured average length is 10.2cm. At the significance level $\alpha = 0.05$, test whether the part length equals 10cm.

Solution:

1. State the hypotheses: $H_0: \mu = 10$, $H_1: \mu \neq 10$
2. Test statistic: $Z = \frac{\overline{X} - 10}{\sqrt{0.04}/\sqrt{16}} = \frac{10.2 - 10}{0.2/4} = 4$
3. Rejection region: $|Z| > 1.96$
4. Decision: $|Z| = 4 > 1.96$, so reject the null hypothesis
5. Conclusion: the part length is not equal to 10cm

When the Variance Is Unknown (T Test)

Population: $X \sim N(\mu, \sigma^2)$, with $\sigma^2$ unknown

Hypotheses:

• $H_0: \mu = \mu_0$
• $H_1: \mu \neq \mu_0$ (two-sided test)
• $H_1: \mu > \mu_0$ (right one-sided test)
• $H_1: \mu < \mu_0$ (left one-sided test)

Test statistic: $T = \frac{\overline{X} - \mu_0}{S/\sqrt{n}} \sim t(n-1)$

Rejection region:

• Two-sided test: $|T| > t_{\alpha/2}(n-1)$
• Right one-sided test: $T > t_{\alpha}(n-1)$
• Left one-sided test: $T < -t_{\alpha}(n-1)$

Example 2: the service life of a brand of batteries follows a normal distribution. 10 batteries of this brand are drawn at random, and the measured average service life is 120 hours with a sample standard deviation of 8 hours. At the significance level $\alpha = 0.05$, test whether the average service life of the batteries is greater than 115 hours.

Solution:

1. State the hypotheses: $H_0: \mu \leq 115$, $H_1: \mu > 115$
2. Test statistic: $T = \frac{120 - 115}{8/\sqrt{10}} = \frac{5}{2.53} = 1.98$
3. Rejection region: $T > t_{0.05}(9) = 1.833$
4. Decision: $T = 1.98 > 1.833$, so reject the null hypothesis
5. Conclusion: the average service life of the batteries is greater than 115 hours

Testing the Variance of a Single Normal Population (Chi-Square Test)

When the Mean Is Known

Population: $X \sim N(\mu, \sigma^2)$, with $\mu$ known

Hypotheses:

• $H_0: \sigma^2 = \sigma_0^2$
• $H_1: \sigma^2 \neq \sigma_0^2$ (two-sided test)
• $H_1: \sigma^2 > \sigma_0^2$ (right one-sided test)
• $H_1: \sigma^2 < \sigma_0^2$ (left one-sided test)

Test statistic: $\chi^2 = \frac{\sum_{i=1}^n (X_i - \mu)^2}{\sigma_0^2} \sim \chi^2(n)$

Rejection region:

• Two-sided test: $\chi^2 < \chi^2_{1-\alpha/2}(n)$ or $\chi^2 > \chi^2_{\alpha/2}(n)$
• Right one-sided test: $\chi^2 > \chi^2_{\alpha}(n)$
• Left one-sided test: $\chi^2 < \chi^2_{1-\alpha}(n)$

When the Mean Is Unknown

Population: $X \sim N(\mu, \sigma^2)$, with $\mu$ unknown

Hypotheses:

• $H_0: \sigma^2 = \sigma_0^2$
• $H_1: \sigma^2 \neq \sigma_0^2$ (two-sided test)
• $H_1: \sigma^2 > \sigma_0^2$ (right one-sided test)
• $H_1: \sigma^2 < \sigma_0^2$ (left one-sided test)

Test statistic: $\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \sim \chi^2(n-1)$

Rejection region:

• Two-sided test: $\chi^2 < \chi^2_{1-\alpha/2}(n-1)$ or $\chi^2 > \chi^2_{\alpha/2}(n-1)$
• Right one-sided test: $\chi^2 > \chi^2_{\alpha}(n-1)$
• Left one-sided test: $\chi^2 < \chi^2_{1-\alpha}(n-1)$

Example 3: the lengths of the parts produced by a factory follow a normal distribution. 10 parts are drawn at random from the products, and the measured sample variance is 0.04. At the significance level $\alpha = 0.05$, test whether the variance of the part lengths equals 0.03.

Solution:

1. State the hypotheses: $H_0: \sigma^2 = 0.03$, $H_1: \sigma^2 \neq 0.03$
2. Test statistic: $\chi^2 = \frac{(10-1) \times 0.04}{0.03} = \frac{9 \times 0.04}{0.03} = 12$
3. Rejection region: $\chi^2 < \chi^2_{0.975}(9) = 2.7$ or $\chi^2 > \chi^2_{0.025}(9) = 19.0$
4. Decision: $2.7 < \chi^2 = 12 < 19.0$, so the null hypothesis is not rejected
5. Conclusion: the variance of the part lengths equals 0.03

Comparison of Testing Methods

Comparison of the Z Test and the T Test

Z test:

• Applicable condition: the variance is known
• Test statistic: $Z \sim N(0,1)$
• Advantages: simple to compute, no need to consult the t distribution table
• Disadvantages: requires knowing the population variance

T test:

• Applicable condition: the variance is unknown
• Test statistic: $T \sim t(n-1)$
• Advantages: does not require knowing the population variance
• Disadvantages: requires consulting the t distribution table

Comparison of Mean Testing and Variance Testing

Mean testing:

• Object of the test: the population mean
• Application scenarios: testing product quality, treatment effects, and so on
• Test statistic: the Z statistic or the T statistic

Variance testing:

• Object of the test: the population variance
• Application scenarios: testing production stability, measurement precision, and so on
• Test statistic: the chi-square statistic

Power Analysis of Tests

Power Function

Factors that influence it:

1. Sample size: the larger the sample size, the higher the power
2. Significance level: the higher the significance level, the higher the power
3. Effect size: the larger the effect size, the higher the power

Determining the Sample Size

Problem: given the significance level, the power, and the effect size, determine the sample size required.

Example 4: for testing the mean of a normal population, given the significance level $\alpha = 0.05$, the power $1 - \beta = 0.8$, and the effect size $d = 0.5$, find the sample size $n$.

Solution:

1. Use the power analysis formula
2. Consult the table or use software to compute
3. We obtain $n \approx 26$

Assumptions of the Tests

The Normality Assumption

Importance: the normality assumption is the basis of the Z test and the T test.

Testing methods:

1. Graphical methods: Q-Q plots, histograms
2. Statistical tests: the Shapiro-Wilk test, the Kolmogorov-Smirnov test

Handling non-normal data:

1. Data transformation: logarithmic transformation, square root transformation, and so on
2. Nonparametric tests: the Wilcoxon signed-rank test, and so on

The Independence Assumption

Importance: the sample observations should be mutually independent.

Testing methods:

1. Time series analysis: the autocorrelation function
2. Spatial data analysis: spatial autocorrelation

Cautions for Testing

The Multiple Comparison Problem

Problem: when performing several hypothesis tests, the probability of committing a Type I error increases.

Solution:

1. Bonferroni correction: divide the significance level by the number of tests
2. Holm correction: adjust the significance level step by step
3. FDR control: control the false discovery rate

Effect Size

Common effect sizes:

1. Cohen’s d: $d = \frac{\mu_1 - \mu_2}{\sigma}$
2. Correlation coefficient: $r = \frac{t}{\sqrt{t^2 + df}}$

Significance: even if the result is statistically significant, a small effect size may mean little practical significance.

Exercises

Exercise 1

Let $X \sim N(\mu, \sigma^2)$ with $\sigma$ known. To test $H_0: \mu=\mu_0$, write the test statistic.

Exercise 2

Let $X \sim N(\mu, \sigma^2)$ with $\sigma$ unknown. To test $H_0: \mu=\mu_0$, write the test statistic.

Exercise 3

Let $X \sim N(\mu, \sigma^2)$ with $\mu$ unknown. To test $H_0: \sigma^2=\sigma_0^2$, write the test statistic.

Exercise 4

Let the sample mean be $\overline{x} = 52$, the population standard deviation be $\sigma = 10$, and the sample size be $n = 25$. Test $H_0: \mu = 50$, $H_1: \mu \neq 50$ at the significance level $\alpha = 0.05$.

Exercise 5

Let the sample mean be $\overline{x} = 120$, the sample standard deviation be $s = 8$, and the sample size be $n = 10$. Test $H_0: \mu \leq 115$, $H_1: \mu > 115$ at the significance level $\alpha = 0.05$.