Chapter 10 Hypothesis Tests

A statistical hypothesis is an assertion or conjecture about a parameter (or parameters) of a population.

For example, the assertion that the mean body temperature of a healthy adult is 98.4 degrees Fahrenheit.

Procedure

To verify such an assertion statistically, one has to

  1. Make a study in which a simple random sample is selected and the value in question is collected for every subject in the sample (in this case the body temperature of the healthy adults).
  2. Set the null hypothesis H and the alternative hypothesis HA.
  3. Find the sample statistics (in this case, the sample mean body temperature, standard deviation, and sample size).
  4. Fix a level of significance (for example 90%, 95% or 98%)
  5. Determine the corresponding critical value.
  6. Calculate the test statistic (see table).
  7. Decide whether the claim is accepted or rejected.

Table for Hypothesis Tests

Hypothesis for parameter conditions test statistic critical values
Proportion p np>=5, nq>=5 z=\frac{\hat{p}-p}{\sqrt{pq/n}} z-score table
Mean (n>=30) \mu \sigma known, n\ge 30 z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} z-score table
Mean (n<30) \mu \sigma known, n< 30 t=\frac{\bar{x}-\mu}{s/\sqrt{n}} t-table, df=n-1
Standard deviation \sigma normally dist. pop. \chi^2=\frac{(n-1)s^2}{\sigma^2} chi-square table, df=n-1 |
Difference of Means (n>=30) \mu_1-\mu_2 \sigma_1,\sigma_2 known, n\ge 30 z=\frac{\bar{x}_2-\bar{x}_2-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+ \frac{\sigma_2^2}{n_2}}} z-score table
Difference of Means (n<30) \mu_1-\mu_2 n< 30 t=\frac{\bar{x}_2-\bar{x}_2-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+ \frac{1}{n_2}}}, where s_p=\sqrt{\frac{(n_1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}} t-table, df=n_1+n_2-2
Difference of Means (paired data) \mu=0 n< 30 t=\frac{\bar{x}-\mu_0}{s/n} t-table, df=n-1

F-distribution

Some statistical tests use the F distribution, which requires two parameters. A table is available here.

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