As a class teacher, your class students took test in both statistics and mathematics. The results are as follows:- mean performance in (i) statistics-52, and (ii) mathematics-48. The students are 20 in number. Standard deviation=16.05, tcal-0.515, ttab=2.121, p=0.05
1 Formulate one null hypothesis and one alternative hypothesis on the students performance in both statistics and mathematics
2. What is the degree of freedom in this situation?
3 state the decision rule to guide you here
4. What decision can make based on your null hypothesis?
5. What is your decision based on your alternative hypothesis?
1. Null hypothesis: There is no significant difference in the performance of students in statistics and mathematics.
Alternative hypothesis: There is a significant difference in the performance of students in statistics and mathematics.
2. The degree of freedom in this situation is (n-1), where n is the number of students. In this case, the degree of freedom would be (20-1) = 19.
3. The decision rule is based on the calculated t-value (tcal) compared to the critical t-value (ttab) at a given significance level (p). If the tcal value falls within the acceptance range of the ttab, we fail to reject the null hypothesis. If the tcal value falls outside the acceptance range, we reject the null hypothesis.
4. Based on the null hypothesis, we would conclude that there is not enough evidence to suggest a significant difference in the performance of students in statistics and mathematics.
5. Based on the alternative hypothesis, we would conclude that there is enough evidence to suggest a significant difference in the performance of students in statistics and mathematics.
1. Null hypothesis (H0): There is no significant difference in the mean performance of the students in statistics and mathematics.
Alternative hypothesis (Ha): There is a significant difference in the mean performance of the students in statistics and mathematics.
2. The degrees of freedom in this situation can be calculated using the formula:
df = (n1 - 1) + (n2 - 1)
where n1 and n2 are the number of students in statistics and mathematics, respectively.
Given that there are 20 students in the class, the degrees of freedom will be df = (20 - 1) + (20 - 1) = 38.
3. The decision rule in this situation is based on the t-distribution critical value.
If the calculated t-value is less than the critical value (tcal < ttab), we fail to reject the null hypothesis.
If the calculated t-value is greater than the critical value (tcal > ttab), we reject the null hypothesis.
4. Based on the null hypothesis, if the calculated t-value is less than the critical value (tcal < ttab), we would conclude that there is no significant difference in the mean performance of the students in statistics and mathematics.
5. Based on the alternative hypothesis, if the calculated t-value is greater than the critical value (tcal > ttab), we would conclude that there is a significant difference in the mean performance of the students in statistics and mathematics.
1. Null Hypothesis (H₀): There is no significant difference in the mean performance between statistics and mathematics for the class.
Alternative Hypothesis (H₁): There is a significant difference in the mean performance between statistics and mathematics for the class.
2. The degrees of freedom in this situation can be calculated as (N₁ + N₂) - 2, where N₁ and N₂ are the sample sizes of statistics and mathematics respectively. Since there are 20 students in total, the degrees of freedom would be (20 + 20) - 2 = 38.
3. The decision rule in this situation would be to compare the calculated t-value (tcal) with the critical t-value (ttab) at a given significance level (α). If the tcal value falls within the critical region, we reject the null hypothesis, otherwise, we fail to reject the null hypothesis. The critical t-value can be obtained from the t-distribution table based on the degrees of freedom and the chosen significance level (p).
4. If we base our decision on the null hypothesis, and the calculated t-value does not fall within the critical region, we would fail to reject the null hypothesis. This would suggest that there is not enough evidence to conclude that there is a significant difference in the mean performance between statistics and mathematics for the class.
5. If we base our decision on the alternative hypothesis, and the calculated t-value falls within the critical region, we would reject the null hypothesis. This would indicate that there is enough evidence to suggest that there is a significant difference in the mean performance between statistics and mathematics for the class.