Hypothesis Testing: Two-tailed test for significance of treatment effect

We are conducting a hypothesis test to determine if the treatment administered to the sample has a significant effect on the sample mean. Here's how we approach the problem:

Step 1: Formulate Hypotheses

• Null Hypothesis (H₀): The treatment has no effect, and the population mean remains at $\mu = 20$.
• Alternative Hypothesis (H₁): The treatment has an effect, and the population mean is different from 20.

This is a two-tailed test because we are looking for any significant deviation from the population mean, regardless of the direction (higher or lower).

Step 2: Calculate the Standard Error of the Mean (SEM)

The standard error of the mean ($\sigma_M$) is given by:

$\sigma_M = \frac{\sigma}{\sqrt{n}}$

Where:

• $\sigma = 10$ (population standard deviation),
• $n$ is the sample size.

Part A: When $n = 25$

For $n = 25$: $\sigma_M = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$

Step 3: Compute the z-score

The z-score is computed as:

$z = \frac{M - \mu}{\sigma_M}$

Where:

• $M = 25$ (sample mean after treatment),
• $\mu = 20$ (population mean before treatment),
• $\sigma_M = 2$ (standard error of the mean).

Substitute the values: $z = \frac{25 - 20}{2} = \frac{5}{2} = 2.5$

Step 4: Determine the Critical z-values

For a two-tailed test with $\alpha = 0.05$, the critical z-values are $\pm 1.96$ (from standard normal distribution tables). If the computed z-score exceeds $\pm 1.96$, we reject the null hypothesis.

• With $z = 2.5$, which is greater than 1.96, we reject the null hypothesis and conclude that the treatment has a significant effect when $n = 25$.

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Part B: When $n = 4$

For $n = 4$: $\sigma_M = \frac{10}{\sqrt{4}} = \frac{10}{2} = 5$

Compute the z-score:

$z = \frac{25 - 20}{5} = \frac{5}{5} = 1$

Step 4: Compare the z-score to critical values

• With $z = 1$, which is less than 1.96, we fail to reject the null hypothesis when $n = 4$. This means that the treatment does not have a statistically significant effect with a smaller sample size.

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Conclusion: Effect of Sample Size

Comparing the two cases:

• With $n = 25$, we found that the treatment had a significant effect ($z = 2.5$).
• With $n = 4$, we found that the treatment did not have a significant effect ($z = 1$).

This demonstrates that increasing the sample size has the following effects:

1. Decreases the value of $\sigma_M$ (standard error of the mean).
2. Increases the value of $z$-score.
3. Increases the likelihood that the hypothesis test will reject the null hypothesis if there is a real effect.

Larger sample sizes make it easier to detect a significant effect, as they reduce variability and make the test more sensitive.

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Would you like more details on any part, or do you have any questions?

5 Related Questions:

1. What is the relationship between sample size and standard error?
2. How do we determine the critical z-values for different confidence levels?
3. Can we conduct a one-tailed test instead, and how would that change the result?
4. How does the effect size influence the likelihood of rejecting the null hypothesis?
5. What are the limitations of using a z-test when the sample size is small?

Tip:

Always check whether the population variance or standard deviation is known before using the z-test. For unknown population variances, the t-test is more appropriate, especially with small sample sizes.