Twenty students randomly assigned to an experimental group receive an instructional program, 30 in the control group do not. After 6 months, both groups are tested 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 a standard deviation of the estimated population of 5). Using the .05 level, what should the experimenter conclude? a) Use the steps of hypothesis testing, b) sketch the distributions involved, c) explain your answer to someone who is familiar with the t test for a single sample, but not with the t test for independent media.
N1 = number of participants in the experimental group
N2 = number of participants in the control group
df1 = degrees of freedom for the experimental group
df2 = degrees of freedom for the control group
dfTotal = degrees of freedom for the two groups of
M1 = mean of experimental group
M2 = mean of the control group
S21 = variance of the estimated population of the USDA Agricultural Research experimental group
S22 = variance of the estimated population of the USDA Agricultural Research control group
S2Pooled = pooled estimate of the population variance
S2M1 = variance of the distribution of means for the experimental group
S2m2 = variance of the distribution of means for the control group
S2Difference = variance of the distribution of differences between means
SDifference = standard deviation of the distribution of the differences between means
t = score for its shows
t = score cutting necessary that sets the rejection region (also known as the critical value)
Decision: Reject the null or fail to reject the null (select only one)
Round your answers to two decimal places.
t = + necessary
#1DukAnswered at 2012-10-26 04:17:28
Is that all the information I gave you? I'm not sure how to do without standard error of the individual scores to compare the predicted scores . In addition , two groups are randomly assigned to the experimental group , but given the exact same variable ? I'm confused about that .
The degree of freedom is the number of steps - 1. It appears that the average known groups. Was there a before and after test to measure , or you're just comparing the results with the control group ?
Estimated variance will be different between the groups because of their size . According to the nominal value of the numbers , I do not think there is a significant difference in the numbers ( at least 95 % to a non CI ) . So do not reject the null hypothesis if this was true .
Unfortunately , this is not much help . Maybe he 's going in the right direction . I'd need each data point from each group to reach the standardized score to do then the t -test (apparently only a single sample t test , twice)
Try to find SPSS , maybe that will give you the numbers you need . The explanation seems easier :)