A random sample is selected from a normal population with a mean of μ = 20 and a standard deviation of σ = 10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 25.
If the sample consists of n = 25 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05.
With n = 25, z =    . You    conclude that the treatment has a significant effect.
Standard Normal Distribution
Mean = 0.0

Standard Deviation = 1.0

-2.0
-1.0
0.0
1.0
2.0
z
.5000
.5000
0.0000
If the sample consists of n = 4 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05.
With n = 4, z =    . You    conclude that the treatment has a significant effect.
Comparing your answers for parts a and b, explain how the size of the sample influences the outcome of a hypothesis test.
Increasing the sample size      the value of σM
M
,      the value of z, and      the likelihood that the hypothesis test will reject the null hypothesis.

### 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 \(n = 25\).

---

#### 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.

---

### 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.

---

### 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.

A random sample is selected from a normal population with a mean of μ = 20 and a standard deviation of σ = 10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 25. If the sample consists of n = 25 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05. With n = 25, z = . You conclude that the treatment has a significant effect. If the sample consists of n = 4 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05. With n = 4, z = . You conclude that the treatment has a significant effect. Comparing your answers for parts a and b, explain how the size of the sample influences the outcome of a hypothesis test. Increasing the sample size decreases the value of σM, increases the value of z, and increases the likelihood that the hypothesis test will reject the null hypothesis.

Explore how to determine if a treatment has a significant effect on a sample mean using a two-tailed z-test. This problem involves calculating the z-score, standard error, and analyzing the effect of sample size. Suitable for students studying introductory statistics.

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