A school experiments with two teaching approaches for mathematics. In one classroom the material is taught using lectures only. In the other classroom lectures are mixed in with the use of manipulatives, computers, and collaborative activities. Do the data below support the claim that students in the classroom with multiple techniques do better. (a=.05)


Lecture​n=20​x=76.0​s=2.1
Multiple​n=32​x=88.5​s=0.4
Help

Number of students in each class?

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

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

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

To determine if the data supports the claim that students in the classroom with multiple techniques do better, we can perform a hypothesis test.

Step 1: State the null and alternative hypotheses.

The null hypothesis (H0) assumes that there is no difference in performance between the two teaching approaches:
H0: μ1 = μ2

The alternative hypothesis (Ha) assumes that students in the classroom with multiple techniques perform better:
Ha: μ1 < μ2

Step 2: Select the significance level.

The significance level (α) is given as 0.05. This means that we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).

Step 3: Compute the test statistic.

Since we are comparing the means of two independent samples, we can use a two-sample t-test. The test statistic is given by:

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

Where:
x1 = mean of the lecture group (76.0)
x2 = mean of the multiple technique group (88.5)
s1 = standard deviation of the lecture group (2.1)
s2 = standard deviation of the multiple technique group (0.4)
n1 = number of students in the lecture group (20)
n2 = number of students in the multiple technique group (32)

Plugging in the values, we get:

t = (76.0 - 88.5) / sqrt((2.1^2 / 20) + (0.4^2 / 32))

Step 4: Determine the critical value.

Since the alternative hypothesis assumes that the students in the multiple technique group perform better, we will perform a one-tailed t-test. We will find the critical value corresponding to a left-tail probability of 0.05 in the t-distribution table for the degrees of freedom given by (n1 - 1) + (n2 - 1).

Step 5: Compare the test statistic with the critical value.

If the test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 6: Interpret the results.

If we reject the null hypothesis, it means that there is enough evidence to support the claim that students in the classroom with multiple techniques do better. If we fail to reject the null hypothesis, it means that there is not enough evidence to support the claim.

Please note that the critical value and the test statistic were not provided in the question. You can find the critical value using a t-distribution table or by using statistical software or an online calculator.