There are two independent groups of students one group took online courses the second group took conventional course face-face. Using the sample data determine if there is a significant difference between the 2 groups scores.Ho(mu1-mu2)=0 (no difference), H1(mu1-mu2)<>0 (there is a difference) a=0.05,t critical=+or-2.01,n1=10,M1=93, s1sq.=20, n2=10, M1=85, s2sq.=20
Here are a few hints:
1. Try an independent groups t-test to calculate the t-test statistic.
2. Degrees of freedom for an independent groups t-test is this:
n1 + n2 - 2
3. Compare your test statistic to the critical values listed. If the test statistic exceeds the critical values, reject the null and conclude a difference. If the test statistic does not exceed the critical values, do not reject the null (you cannot conclude a difference in this case).
I hope this will get you started.
To determine if there is a significant difference between the scores of the two groups, we can conduct a two-sample t-test.
Step 1: State the hypotheses
Ho (null hypothesis): There is no significant difference between the means of the two groups.
H1 (alternative hypothesis): There is a significant difference between the means of the two groups.
Ho: mu1 - mu2 = 0
H1: mu1 - mu2 ≠ 0
Step 2: Determine the significance level (alpha)
The significance level, denoted by "a", is given as 0.05. This means we want to be 95% confident in our results.
Step 3: Calculate the critical t-value
The critical t-value is determined by the degrees of freedom (df) and the significance level. In this case, the degrees of freedom would be (n1 - 1) + (n2 - 1), where n1 and n2 are the sample sizes of the groups. The critical t-value for a two-tailed test at a 0.05 significance level with 18 degrees of freedom (10+10-2 = 18) is approximately ±2.01.
Step 4: Calculate the test statistic
The test statistic is calculated using the formula:
t = (M1 - M2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where M1 and M2 are the means, s1^2 and s2^2 are the variances, and n1 and n2 are the sample sizes of the two groups.
Given the following values:
n1 = 10, M1 = 93, s1^2 = 20
n2 = 10, M2 = 85, s2^2 = 20
t = (93 - 85) / sqrt((20/10) + (20/10))
= 8 / sqrt(2 + 2)
= 8 / sqrt(4)
= 8 / 2
= 4
Step 5: Compare the test statistic with the critical value
Since the absolute value of the calculated test statistic (4) is greater than the critical t-value (+/-2.01), we reject the null hypothesis. This means there is a significant difference between the means of the two groups.
In conclusion, based on the sample data, we can conclude that there is a significant difference between the scores of the two groups.