version of the course that involved significant computer interaction through the
ID: 3290036 • Letter: V
Question
version of the course that involved significant computer interaction through the use of statistical packages. Group 2 was chosen from among those who had taken the version of the course that did not use computers. The two groups were from independent populations. The students' attitudes were measured by administering the Computer Comfort Index (CC1). The results were as follows: Group 1 (with computers): n = 15, mean = 67.2, s = 2.1 Group 2 (without computers): n = 10, mean = 60.3, s = 7.5 Test a claim that the mean computer comfort score of those completing the version of the course with computer experience is significantly greater than those completing it without computer experience at significance level of 0.05. (Assume normal distributions). [Claim: u (with computer - without computer) > 0 (using 05 level of significance)].Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1< 2 or 1 - 2< 0
Alternative hypothesis: 1 > 2 or 1 - 2 > 0 (Claim)
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(2.12/15) + (7.52/10)] = 2.4328995047
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (2.12/15 + 7.52/10)2 / { [ (2.12 / 15)2 / (14) ] + [ (7.52 / 10)2 / (9) ] }
DF = 35.03456 / (0.006174 + 3.515625) = 9.947 or 10
t = [ (x1 - x2) - d ] / SE = [ (67.2 - 60.3) - 0 ] / 2.4328995047 = 2.84
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 10 degrees of freedom is more extreme than 2.84.
We use the t Distribution Calculator to find P(t < 2.84)
The P-Value is 0.008774.
The result is significant at p < 0.05.
Interpret results. Since the P-value (0.008774) is less than the significance level (0.05), we cannot accept the null hypothesis.
Conclusion. Reject the null hypothesis. We have sufficient evidence to prove the claim.
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