**Study on Effectiveness of Remedial Tutoring**A statistics tutor aims to evaluate the effectiveness of her remedial tutoring for 10 students. The assessment employs a pre-post design, recording grades before and after the tutoring sessions. The table below presents the students' scores:---**Tasks:**i. **Hypothesis Testing Using SPSS:** Conduct a statistical test to determine the effectiveness of the tutoring at a 0.05 significance level.ii. **Test Statistic Value & Conclusion:** Report the value of the test statistic and conclude whether the null hypothesis is retained or rejected.iii. **Effect Size Calculation:** Compute the effect size using estimated Cohen’s d.---**Scores:**| Before | After ||--------|-------|| 3.9 | 2.2 || 3.1 | 4.8 || 2.7 | 3.9 || 2.4 | 2.9 || 4.3 | 3.8 || 4.4 | 4.4 || 1.1 | 2.7 || 1.9 | 2.2 || 2.8 | 3.9 || 2.9 | 1.5 |---Utilize this data in SPSS to perform the analysis and interpret the results to understand the impact of tutoring.

BeforeAfter3.92.23.14.82.73.92.42.94.33.84.444.12.71.92.22.83.92.91.5i) Null hypothesis: H0:μDifference=0 Alternate hypothesis: HA:μDifference≠0 This is two tailed test. The significance level is 0.05 and we will conduct a matched-pairs t-test of the null hypothesis. Use the sample data to calculate the standard deviation of the differences(s), the standard error (S.E) of the mean difference, the degrees of freedom(DF), and the t statistic test statistic(t). StudentBeforeAfterDifference(d)13.92.21.72.689623.14.8-1.73.097632.73.9-1.21.587642.42.9-0.50.313654.33.80.50.193664.440.40.115674.12.71.41.795681.92.2-0.30.129692.83.9-1.11.3456102.91.51.41.7956Sum32.531.90.613.064Mean3.253.190.061.3064ii) The P-value for this test is the probability that a t statistic having 9 degrees of freedom is less than - 0.16 or greater than 0.16. P-value = P(t < -0.16) + P(t > 0.16) Use the excel formula, "=T.DIST.2T(0.16,9)" to determine the P-value. P-value = 0.8764Decision: Since the P-value (0.8764) is greater than the significance level (0.05), hence we failed to reject the null hypothesis. Conclusion: From the above test we do not have sufficient evidence in the favor of the claim that her tutoring is effective.iii)The effect size using estimated Cohen's d is 0.05. The observed effect size d is very small, 0.05. SPSS steps:Step 1) Launch SPSS and open your dataset.Step 2) From the menu bar, go to "Analyse" and select "Compare Means" in the drop-down menu.Step 3) Click on "Paired-Samples T Test" in the submenu.Step 4) Move the variable you want to test from the list of available variables to the "Paired Variables" box.Step 5) Define Hypotheses:Null Hypothesis(H0): μD=0Alternative Hypothesis(HA): μD≠0Step 6) Click "OK" to run the analysis.Output:i) Null hypothesis: H0:μDifference=0 Alternate hypothesis: HA:μDifference≠0 This is two tailed test. The significance level is 0.05 and we will conduct a matched-pairs t-test of the null hypothesis. ii)t = 0.16P-value = 0.8764Decision: Since the P-value (0.8764) is greater than the significance level (0.05), hence we failed to reject the null hypothesis. Conclusion: From the above test we do not have sufficient evidence in the favor of the claim that her tutoring is effective.iii) The effect size using estimated Cohen's d is 0.05. The observed effect size d is very small, 0.05.SPSS steps:Step 1) Launch SPSS and open your dataset.Step 2) From the menu bar, go to "Analyse" and select "Compare Means" in the drop-down menu.Step 3) Click on "Paired-Samples T Test" in the submenu.Step 4) Move the variable you want to test from the list of available variables to the "Paired Variables" box.Step 5) Define Hypotheses:Null Hypothesis(H0): μD=0Alternative Hypothesis(HA): μD≠0Step 6) Click "OK" to run the analysis.Output: