1. Hypothesis Testing: Hypotheses
The null hypothesis is usually specified so that the null parameter is equal to a specific value, or that two parameters are equivalent.
Example:
𝐻 0 : 𝜇= 𝜇 0 =5 ,
𝐻 0 : 𝜇 1 = 𝜇 2
We will often call the null hypothesis, the hypothesis of no effect.
The alternative hypothesis may be either
One-tailed (or, one-sided): The alternative specifies a region either to the left (<) or right (>) of the null parameter.
E.g., 𝐻 1 : 𝜇> 𝜇 0
Two-tailed (or, two-sided, or non-directional): The alternative specifies a region to both the left or right of the null parameter.
E.g., 𝐻 1 : 𝜇≠ 𝜇 0

2. Decision Rules and Errors
Decision rules are a set of guidelines that we use to determine (based on our gathered sample data) whether we should accept or reject 𝐻 0 .
Type I Error: 𝑃 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 0 𝐻 0 𝑡𝑟𝑢𝑒 =𝛼 (also called a false positive)
Type II Error: 𝑃 𝑎𝑐𝑐𝑒𝑝𝑡 𝐻 0 𝐻 1 𝑡𝑟𝑢𝑒 =𝛽 (also called a false negative)
Power: 𝑃 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 0 𝐻 1 𝑡𝑟𝑢𝑒 =1−𝛽 (also called a true positive)
Correct negative: 𝑃 𝑎𝑐𝑐𝑒𝑝𝑡 𝐻 0 𝐻 0 𝑡𝑟𝑢𝑒 =1−𝛼
For hypothesis testing, only Type I error can be set manually.
By convention, it is common to set it as 𝛼=0.05 or 0.01.
One should really try to find an appropriate balance between Type I and Type II errors, but this is complicated.
Type II error is a function of effect size, sample size, and Type I error.
𝑯 𝟎 True
𝑯 𝟏 True
Reject 𝑯 𝟎
Type I error (𝛼)
Power (1−𝛽)
Accept 𝑯 𝟎
Correct (1−𝛼)
Type II error (𝛽)

3. Critical Values
A critical value is the value 𝑧 𝛼 such that if our observed z statistic is “more extreme” than 𝑧 𝛼 we reject 𝐻 0 .
The region greater than (if 𝑧 𝛼 > 𝑧 𝑜𝑏𝑠 ) or less than (if 𝑧 𝛼 < 𝑧 𝑜𝑏𝑠 ) the critical value is called the critical region.
This region is determined by the value of 𝛼, as well as the null hypothesis.
As mentioned, the convention is to set 𝛼=0.05 or 0.01.
We find the critical value 𝑧 𝛼 that corresponds to this 𝛼-level.

4. Critical Values
Suppose the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇≠ 𝜇 0 .
𝜇 is the true mean.
𝜇 0 is the hypothesized value for the true mean.
Since this is two-sided, there are two critical values: ± 𝑧 𝛼 2 .
Suppose the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇< 𝜇 0 .
There is only one critical value: − 𝑧 𝛼 .
Suppose the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇> 𝜇 0 .
There is only one critical value: 𝑧 𝛼 .

5. p-values
A p-value is the probability that a test statistic takes on a value as extreme or more extreme (in the direction of the alternative hypothesis) than the test statistic’s observed value, assuming 𝐻 0 is true.
Say, the value of the test statistic is z. (The random variable of the test statistic is Z).
If the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇< 𝜇 0 , then
𝑝=𝑝 𝑍≤𝑧
If the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇> 𝜇 0 , then
𝑝=𝑝 𝑍≥𝑧
If the hypotheses are 𝐻 0 :𝜇= 𝜇 0 vs. 𝐻 1 :𝜇≠ 𝜇 0 , then
𝑝=𝑝 𝑍≥|𝑧| =𝑝 𝑍≤−𝑧 +𝑝(𝑍≥𝑧)
If it is further supposed that the distribution of Z is symmetric, then
𝑝=𝑝 𝑍≥|𝑧| =𝑝 𝑍≤−𝑧 +𝑝 𝑍≥𝑧 =2𝑝 𝑍≥𝑧
A small p-value means that the probability of obtaining your observed test statistic (assuming 𝐻 0 is true) is small.
Thus, the smaller the p-value, the more evidence have against 𝐻 0 .
Show examples of the three hypotheses on the board. Put the distribution, show alpha, z_alpha (or z_{alpha/2}).

6. Conducting Hypothesis Tests
For conducting hypothesis tests for the population mean, we will use the sample mean statistic.
Either normality will be assumed, or we can make use of the CLT (assuming the sample size is large enough).
For now, we will also assume the population variance is known.
When the population variance is not known, then it must be estimated, which will change the distribution of the test statistic. This will be further explored later.

7. Example
A developmental psychologist claims that a training program he developed according to a theory should improve problem- solving ability. For a population of 7-year-olds, the mean score 𝜇 on a standard problem-solving test is known to be 80 with a standard deviation of 10. To test the training program, 26 7-year-olds are selected at random, and their mean score is found to be 82. Let’s assume the population of scores is normally distributed. Can we conclude, at an 𝛼=.05 level of significance, that the program works? Assume the sd of scores after the training program is also 10.
We hypothesize that the test improves problem-solving. What are the null and alternative hypotheses?
𝐻 0 :𝜇 = 80 vs. 𝐻 1 :𝜇>80
The data are normally distributed, so 𝑋 ~𝑁 𝜇, 𝜎 2 𝑛
A z-statistic can then be formed:
𝑧= 𝑥 − 𝜇 0 𝜎 𝑛 = 82− ≈1.0198
p-value method:
𝑝=𝑝 𝑍≥ =.15>.05
Since 𝑝>𝛼, then we cannot reject 𝐻 0 .
Critical value method:
𝛼=.05→ 𝑧 𝛼 =
Since 𝑧=1.0198< = 𝑧 𝛼 , then we cannot reject 𝐻 0 .
Substantively, this means that a mean score of 82 is likely to occur by chance, so we cannot say that the training program improved problem-solving skills in 7-year-olds.
Show on the board the duality of using p-values and using the critical value method.