One Sample z-test and Hypothesis Testing

The one sample z-test is a specific type of hypothesis test used to determine if a treatment has a significant effect on a sample compared to a known population.
In this context, a z-score is used because the population standard deviation is typically known, making it a natural starting point for learning hypothesis testing.
Experimental Group vs. Control Group:
- In a classic experiment, there are two groups: an experimental group (which receives a treatment like a pill, exercise, or therapy) and a control group (which does not receive the treatment).
- Comparison between these two groups allows researchers to determine if the treatment was effective (e.g., if Tylenol reduces headache pain to a level of 3 while the control group remains at a 6).
One Sample Test Specifics:
- This specific test involves only one group (one sample) that receives the treatment.
- There is no second sample or separate control group.
- Instead, the sample is compared to the general population, which effectively serves as the control group.
- This method is only appropriate when the parameters (mean and standard deviation) of the general population are already known.

Hypothesis tests can be categorized based on the specificity of the prediction.
One-Tailed Test:
- Used when there is a specific directional prediction.
- Examples: Predicting a pill will only decrease pain, a therapy will only decrease depression, or a memory pill will only increase memory.
Two-Tailed Test:
- Used for a general prediction where the treatment might increase or decrease the variable, or when the researcher simply wants to see if the treatment has any effect at all.
- This is more common in early experimental stages (e.g., a new drug tested on rats) where the outcome is uncertain.
- It is often the default choice to avoid making a directional prediction that might be proven wrong (e.g., predicting an increase in memory when the drug actually causes a decrease).
Alpha Levels ($\alpha$):
- The Alpha level is the cutoff for significance.
- Standard Cutoff: $\alpha = 0.05$
- Extra Impressive Cutoff: $\alpha = 0.01$

Scenario: A researcher wants to know if a newly discovered antidepressant called Tinocell (also referred to phonetically as Temisil, Temocell, or Tebusill) affects depression symptoms.
The Population: Because testing the entire population of millions of depressed patients is impossible, a sample is used.
The Sample:
- Size ( $n$ ): $30$ clinically depressed patients.
- Treatment: Participants take Tinocell for four weeks.
- Result: Sample mean ( $M$ ) = $2.2$ depression symptoms per week.
The Known Population Parameters (General population not taking the drug):
- Population Mean ( $\mu$ ): $3.2$ symptoms per week.
- Population Standard Deviation ( $\sigma$ ): $3$ .
Test Requirements:
- Use a two-tailed test.
- Use an alpha level of $\alpha = 0.05$ .

Hypotheses must be stated in both words and symbols.
Null Hypothesis ( $H_0$ ):
- Often associated with the idea of "no difference, no preference, or no effect."
- In words: "Tinocell does not affect depression" or "Tinocell has no effect on depression."
- In symbols: $H_0: \mu = 3.2$ .
- This assumes that if you gave the drug to the whole population, the average symptoms would remain the same as the general population.
Alternative Hypothesis ( $H_1$ ):
- The opposite of the null hypothesis.
- In words: "Tinocell has an effect on depression."
- In symbols: $H_1: \mu \neq 3.2$ .
- This indicates the drug changes the mean, though it doesn't specify if it increases or decreases it in a two-tailed test.
Note on Symbols: Hypotheses are always about the population mean ( $\mu$ ), never the sample mean ( $M$ ). We already know the sample mean ( $M = 2.2$ ); the goal is to make an inference about the population.

Distribution of Sample Means: Draw a normal bell curve representing the distribution as if the null hypothesis ( $H_0$ ) were true.
Center of the Distribution:
- Mark the center as the population mean from the null hypothesis: $\mu = 3.2$ .
- At the center, the z-score is always zero ( $z = 0$ ).
Critical z-scores ( $z_{crit}$ ):
- Based on $\alpha = 0.05$ and a two-tailed test, the 5% alpha is split between the two tails (2.5% in each tail).
- Using the unit normal table, these cutoff values are found to be: $z = \pm 1.96$ .
Critical Regions: The areas in the tails past $+1.96$ and $-1.96$ are shaded. If the calculated $z_{obtained}$ falls in these shaded regions, the null hypothesis is rejected.

Decision Rule: Reject $H_0$ if the $z_{obtained}$ is in the critical region (the shaded tails). Otherwise, fail to reject (retain) $H_0$ .
Evaluation:
- The obtained z-score is $-1.83$ .
- The critical boundary on the negative side is $-1.96$ .
- On a number line, $-1.83$ is to the right of $-1.96$ ; it is not far enough out to fall into the shaded tail.
Statistical Decision: Fail to Reject $H_0$ (or Retain $H_0$ ).
Real-World Conclusion:
- State the result in plain language that avoids statistical jargon like "null" or "alpha."
- Conclusion: "Tinocell has no effect on depression" or "Tinocell does not significantly affect depression."
Final Note: If the obtained value had been more extreme (e.g., $z = -2.1$ ), the null hypothesis would have been rejected, allowing the researcher to conclude that the drug does have an effect.