Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude?

(a) and (c)

Ho: u(ex) - u(ct) = 0
Ha:u(ex) – u(ct) is not equal to 0
Test stat: t = (38-35) / sqrt[9/20 + 25/30] = 2.6482
p-value = 2*P(t > 2.6482 with df = 48) = 0.0109
Conclusion:
At the 5% significance level, reject Ho because the p-value is less than 5%

To determine what the experimenter should conclude, we need to perform a hypothesis test. We will compare the means of the experimental group and the control group.

Step 1: State the hypotheses:
Null hypothesis (H₀): The instructional program has no effect on the knowledge of the students. The population means are equal.
Alternative hypothesis (H₁): The instructional program has an effect on the knowledge of the students. The population means are not equal.

Step 2: Set the significance level:
The significance level, also denoted as α, is given as .05. This means we want to be 95% confident in our conclusion.

Step 3: Calculate the test statistic:
Since the sample sizes are relatively small (less than 30), we should use the t-distribution. The formula for the t-test is:

t = (x₁ - x₂) / sqrt((s₁²/n₁) + (s₂²/n₂))

Where:
x₁ = mean of the experimental group = 38
x₂ = mean of the control group = 35
s₁ = estimated population standard deviation of the experimental group = 3
s₂ = estimated population standard deviation of the control group = 5
n₁ = number of students in the experimental group = 20
n₂ = number of students in the control group = 30

Let's calculate the test statistic:

t = (38 - 35) / sqrt((3²/20) + (5²/30))

Simplifying the equation:

t = 3 / sqrt(0.45 + 0.2778)
t = 3 / sqrt(0.7278)
t ≈ 3 / 0.8537
t ≈ 3.514

Step 4: Determine the critical t-value:
Since the sample sizes are small, we need to look up the critical t-value from the t-distribution table. The degrees of freedom (df) are calculated as:

df = (n₁ + n₂) - 2 = (20 + 30) - 2 = 48

Using a significance level of .05 with a two-tailed test, the critical t-value is approximately ±2.013.

Step 5: Make a decision:
Based on the calculated test statistic (3.514) and the critical t-value (±2.013), we compare the absolute value of the test statistic to the critical value.

Since the absolute value of the test statistic (3.514) is greater than the critical value (2.013), we can reject the null hypothesis (H₀).

Step 6: State the conclusion:
We have enough evidence to conclude that the instructional program has a statistically significant effect on the students' knowledge. Therefore, the experimenter should conclude that the instructional program has a positive impact on the students' knowledge after 6 months.

To determine what the experimenter should conclude, we can perform a hypothesis test. The goal is to compare the means of the experimental group and the control group to see if there is a significant difference.

Let's go through the steps of conducting this hypothesis test:

Step 1: State the null hypothesis (H0) and alternative hypothesis (H1):
- The null hypothesis states that there is no significant difference between the means of the two groups. In this case, H0: μ1 = μ2 (where μ1 is the mean of the experimental group and μ2 is the mean of the control group).
- The alternative hypothesis states that there is a significant difference between the means of the two groups. In this case, H1: μ1 ≠ μ2.

Step 2: Determine the level of significance (α):
- The level of significance, denoted by α, is the probability of rejecting the null hypothesis when it is true. In this case, the significance level is given as .05.

Step 3: Compute the test statistic:
- We can use the independent samples t-test to compare the means of the two groups. The formula for the test statistic is:

t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))

Where:
x1 = mean of the experimental group
x2 = mean of the control group
s1 = estimated population standard deviation of the experimental group
s2 = estimated population standard deviation of the control group
n1 = number of observations in the experimental group
n2 = number of observations in the control group

Plugging in the given values:
x1 = 38, x2 = 35, s1 = 3, s2 = 5, n1 = 20, n2 = 30

Step 4: Determine the critical value:
- The critical value(s) depend on the level of significance (α) and the degrees of freedom. In this case, since we are using a two-tail test (H1: μ1 ≠ μ2), we need to divide the significance level by 2 (.05 / 2 = .025) and find the critical value(s) in the t-distribution table with the appropriate degrees of freedom.

Step 5: Make a decision:
- If the test statistic falls within the critical region (beyond the critical value(s)), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, without knowing the degrees of freedom, we can't calculate the exact critical value. However, we can use statistical software or consult a t-distribution table to find it.

With the test statistic and critical value(s), we can compare them. If the test statistic falls beyond the critical value(s), we reject the null hypothesis. If it falls within the critical value(s), we fail to reject the null hypothesis.

By following these steps, the experimenter can conclude whether there is a significant difference between the means of the experimental and control groups.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.