With the data collected, Dr Tian ran a test to see if there was a significant difference in the mean of X (a factor affecting the continued usage of e-learning communication) between male and female students. Given, the data set is: For male: n1 = 15x⎯⎯1 = 4.66s1 = 1.676n1 = 15x¯1 = 4.66s1 = 1.676 For Female, n2 = 15x⎯⎯2 = 3.63s2 = 1.008 Independent sample t-test is used because samples are independent. Both are independent variaables. Null and Alternative hypothesis : H0: μ1 = μ2VsHa: μ1 ≠ μ2H0: μ1 = μ2VsHa: μ1 ≠ μ2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. First find test statistics: Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t = x⎯⎯1 − x⎯⎯2(n1−1)s21 + (n2−1)s22(n1 + n2 − 2) (1n1 + 1n2)√t = 4.66 − 3.63(15−1) × 1.6762 + (15−1) × 1.0082(15 +15 −2)× (115 + 115)√t = 2.04t = x¯1 - x¯2(n1-1)s12 + (n2-1)s22(n1 + n2 - 2) 1n1 + 1n2t = 4.66 - 3.63(15-1) × 1.6762 + (15-1) × 1.0082(15 +15 -2)× 115 + 115t = 2.04 p-value calculation: df = n1 + n2 - 2 = 28 significance level = 0.05 and test is two -tailed test. By using t-calculator the p-value is: p-value= 0.0509 Decision : P-value (0.0509) > significance level (0.05), it fails to reject the null hypothesis. Reject the null hypothesis. Conclusion: There is not significant evidence that there is significant gender difference. Question: Decision : P-value (0.0509) > significance level (0.05), it fails to reject the null hypothesis. Reject the null hypothesis. Conclusion: There is not significant evidence that there is significant gender difference. Question: Based on the result and the choice of your X, suggest whether Dr Tian should use two different ways to motivate male and female students to use e-learning communication, respectively.
With the data collected, Dr Tian ran a test to see if there was a significant difference in the mean of X (a factor affecting the continued usage of e-learning communication) between male and female students.
Given,
the data set is:
For male:
n1 = 15x⎯⎯1 = 4.66s1 = 1.676n1 = 15x¯1 = 4.66s1 = 1.676
For Female,
n2 = 15x⎯⎯2 = 3.63s2 = 1.008
Independent sample t-test is used because samples are independent.
Both are independent variaables.
Null and Alternative hypothesis :
H0: μ1 = μ2VsHa: μ1 ≠ μ2H0: μ1 = μ2VsHa: μ1 ≠ μ2
This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
First find test statistics:
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:
t = x⎯⎯1 − x⎯⎯2(n1−1)s21 + (n2−1)s22(n1 + n2 − 2) (1n1 + 1n2)√t = 4.66 − 3.63(15−1) × 1.6762 + (15−1) × 1.0082(15 +15 −2)× (115 + 115)√t = 2.04t = x¯1 - x¯2(n1-1)s12 + (n2-1)s22(n1 + n2 - 2) 1n1 + 1n2t = 4.66 - 3.63(15-1) × 1.6762 + (15-1) × 1.0082(15 +15 -2)× 115 + 115t = 2.04
p-value calculation:
df = n1 + n2 - 2 = 28
significance level = 0.05
and test is two -tailed test.
By using t-calculator the p-value is:
p-value= 0.0509
Decision :
P-value (0.0509) > significance level (0.05), it fails to reject the null hypothesis.
Reject the null hypothesis.
Conclusion:
There is not significant evidence that there is significant gender difference.
Question:
Decision :
P-value (0.0509) > significance level (0.05), it fails to reject the null hypothesis.
Reject the null hypothesis.
Conclusion:
There is not significant evidence that there is significant gender difference.
Question: Based on the result and the choice of your X, suggest whether Dr Tian should use two different ways to motivate male and female students to use e-learning communication, respectively.
![Independent Samples Teste
Levene's
Test for
Equality of
Variancese
t-test for Equality of Meanse
95% Confidence
Interval of the
Sig. (2-
Mean
Std. Error
Difference
Fe Sig.
dfe
tailed)e Difference Difference Lowere Upper
te
Preference Equal
Before
variances
5.63
.Ow 2.046
28
.049
1.033
.505
-.001
2.068
Visitinge
assumede
Equal
2.046- 22.959
.052
1.033
505
-.012
2.078
variances not
assumed
[Important Note: w=the 7th digit of your student ID. E.g. s1234576. So, w=6]-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86f1854c-b774-4b9e-8d8d-e19f0c605752%2F4ac4597c-1c63-4529-98a0-1b6d2927d220%2Fy1leq5a_processed.png&w=3840&q=75)
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