Need Help with t-test for independent means Please!

[chianti72116]

Hi All,
I am having a heck of a time with this statistics assignment! Just when I think I understand what the assignment is asking for, I am completely wrong and I have to start over again. I have asked for clarification from my instructor, but she just seems to be confusing me even more! I have filled in what I could figure out as best I could, the rest, I REALLY need help with! If someone could show/explain to me what I am supposed to be doing, I would be eternally grateful


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) Use the steps of hypothesis testing for a t-test for independent means

Step 1 In other words… Your answer
Restate the question as: -- --
a. a research hypothesis Research hypothesis = Research hypothesis is that students, who received the instructional program after 6 months, will score higher when tested on their knowledge.
b. a null hypothesis Null hypothesis = Null hypothesis is that both groups of students will score about the same when tested on their knowledge.

Step 2 Determine the characteristics of the comparison distribution -- --
a. The mean will be 0 Mean = 0
b. Find the standard deviation -- --
1. Figure the estimated population variances based on each sample
(For each population, the estimated population variance = the squared deviation from the sample’s mean/degrees of freedom) Because the standard deviations are given (and are the square root of the estimated variances), the estimated variances are found by squaring the standard deviations given. Therefore,

The estimated population variance based on the experimental group = 9

The estimated population variance based on the control group = 25

2. Figure the pooled estimate of the population variance (Degrees of freedom for the experimental group/the total degrees of freedom)(the estimated population variance for the experimental group) + (Degrees of freedom for the control group/the total degrees of freedom)(the estimated population variance for the control group) 17
3. Figure the variance of each distribution of means. The variance of the experimental group distribution of means is: the pooled estimate of the population variance/number of subjects in the experimental group = ?

The variance of the control group distribution of means is: the pooled estimate of the population variance/number of subjects in the control group =
4. Figure the variance of the distribution of differences between the means The variance of the experimental group distribution of means + the variance of the control group distribution of means =
5. Figure the standard deviation of the distribution of differences between means The square root of the variance of the distribution of differences between the means
Step 3 Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected -- --
a. Determine the total degrees of freedom, the significance level and whether to use a one-tailed or a two-tailed test The total degrees of freedom =
The significance level =
The number of tails =
b. Lookup the appropriate cutoff in a t table The cutoff =
Step 4 Determine your sample’s score on the comparison distribution (Mean of the experimental group- mean of the control group)/standard deviation of the distribution of differences between means (hint: it should be between 2 and 3) =
Step 5 Decide whether to reject the null hypothesis Compare the results from step 3 and 4 to accept or reject the null hypothesis

 

[royhaas]

In order to use the t-test, you must assume that the two population variances are equal. If \(\displaystyle S^2\) is the pooled variance, then the extimated variance of the mean difference is simply \(\displaystyle S^2(1/n_1 + 1/n_2)\).