Problem 1: Probability Using Standard Variable z and Normal Distribution Tables
The problem at hand is that there is a need to determine which of the therapies administered is effective in the management of ADHD. Ten children received the ten different interventions each. They were given science textbook to read and their reading time was measured in seconds. The number of seconds were then compared to the standard mean. The z-score and p-value can help researchers determine which of the therapies is statistically significant. In the first test, the standard deviation of 30 was used.
Hypothesis
- Null hypothesis: There is no significant difference between the experimental mean and the population mean at 95% confidence level in all the interventions conducted.
- Alternative Hypothesis: The experimental mean and the population mean is different at 95% confidence level.
Condition
If the p-value is less than the alpha value of 0.05, the researchers reject the null hypothesis and conclude that there is a significant difference between the experimental mean and population mean in all the interventions conducted.
Findings
After running the tests using the standardized z and normal distribution table, the following results were obtained. The p-value of each of the intervention is recorded in the last column as shown in the table below.
µ = 100 seconds and σ = 30
Conclusion
Which child or children, if any, appeared to come from a significantly different population than the one used in the null hypothesis?
From the results, it is clear that out of the ten interventions conducted, none of them is statistically significant. All of the interventions have p-value of more than 0.05. Consequently, it is possible to conclude that all of the ten interventions administered are not effective in changing the reading speed of the child. In other words, the interventions administered are not effective enough to improve the reading attention of children with ADHD.
What happens to the “significance” of each child’s data as the data are progressively more dispersed?
Since the p-value is less than the alpha value of 0.05, it is possible to conclude by saying that the average time taken by the ten children reading after the therapy is not statistically different meaning that the therapy did not improve their reading attention.
APA statement of conclusion
For each of the ten children, the therapy they received was not effective enough to improve their reading attention time as evidenced by the fact that their mean reading after therapy is significantly different from their reading duration before the therapy.
- Child 1. Since the p-value calculated (0.203) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 2. Since the p-value calculated (0.264) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 3. Since the p-value calculated (0.356) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 4. Since the p-value calculated (0.488) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 5. Since the p-value calculated (0.309) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 6. Since the p-value calculated (0.184) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 7. Since the p-value calculated (0.102) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 8. Since the p-value calculated (0.097) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 9. Since the p-value calculated (0.081) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 10. Since the p-value calculated (0.055) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
To determine whether the variation between the samples had some influence on z and p-value, similar experiment was analyzed using the standard deviation of 40. The findings are summarized on table below
µ = 100 seconds and σ = 40
Which child or children, if any, appeared to come from a significantly different population than the one used in the null hypothesis?
Although the data are increasingly more spread compared to table 3 data, none of ten therapies administered was effective in changing the reading speed of the children. In other words, the mean reading speed of the children before and after the therapy is statistically the same. Similar to the first study depicted in table 3, no child benefited from intervention. In fact, the only difference is that the significance difference widen.
What happens to the “significance” of each child’s data as the data are progressively more dispersed?
By making the data more distributed, the p-value move further from 0.05 meaning that the mean reading time after intervention is almost similar to the mean reading time before intervention.
APA statement of conclusion
The more dispersed the data, the lesser the effectiveness of the intervention becomes.
- Child 1. Since the p-value calculated (0.264) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 2. Since the p-value calculated (0.316) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 3. Since the p-value calculated (0.390) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 4. Since the p-value calculated (0.488) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 5. Since the p-value calculated (0.352) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 6. Since the p-value calculated (0.248) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 7. Since the p-value calculated (0.171) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 8. Since the p-value calculated (0.164) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 9. Since the p-value calculated (0.147) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
- Child 10. Since the p-value calculated (0.115) is more than the alpha value set (0.05), we fail to reject the null hypothesis and conclude that there is no significant difference between the mean reading time before and after the therapy.
Problem 2: Two-Sample Inferences
In problem 2, the issue that was being investigated was whether the TB vaccination was effective or not. To address the issue, a group comprising of TB cases from 10 geographical regions received the TB vaccines. The patients in each region were then followed up to see whether new cases would increase or decrease. The data was recorded as shown in table below.
Hypothesis
- The null hypothesis is that the mean of the cases before vaccination and the cases after vaccination sets of arithmetic tests are equal.
- Therefore, the alternative hypothesis is that mean of the cases before vaccination and the cases after vaccination sets of arithmetic tests are not equal.
Set significance level
= 0.05.
Determine the appropriate test statistic
The most appropriate statistical test adopted is the paired t-test. The paired t-test (test of the difference in mean value) is to determine whether there is a difference between two dependent samples (Rosner, 2016). The minitab program was used to perform the analysis.
Findings
The findings from the paired t-test generated is shown below.
Conclusion
Since the p-value is less than 0.005, we reject the null hypothesis and conclude that there is significant difference between the average scores of the cases before vaccination and the cases after vaccination. The findings show that the TB vaccination is effective because the regions which received the vaccination registered a significantly lower number of new cases. The implication of the findings is that the researchers can now recommend the need for vaccination because it is apparent that vaccination is likely to reduce the number of new TB cases.
Problem 3: Cross-Sectional Study
The last problem involves cases of poisoning with chemicals in the homes of 100,000 people in two regions. The samples are independent because the way the data was collected. The issue is to determine whether the cases of poisoning in the two regions is statistically significant or not.
Hypothesis
- The null hypothesis is that the mean of the poisoning cases in region 1 is equal to the mean of the poisoning cases in region 2.
- Therefore, the alternative hypothesis is the mean of poisoning cases in region 1 is not equal to the mean of poisoning cases in region 2.
Set significance level
= 0.05.
Determine the appropriate test statistic
The most appropriate statistical test adopted is the unpaired t-test. The unpaired t-test (test of the difference in mean value) is to determine whether there is a difference between two independent samples (Rosner, 2016). The minitab program was used to perform the analysis.
Findings
The findings from the chart generated is shown in table below
Conclusion
From the table, it is clear that the p-vale is less than 0.05. Since the p-value calculated is less than the alpha value, we reject the null hypothesis and conclude that there is significant difference in the number of poisoning cases between the two regions.
The implication of the findings is that one of the regions could be having support measures to prevent its citizens from engaging in poisoning. There are many other factors that can explain the difference. It is, therefore, crucial for the public health care team to do more research to establish what could be the underlying factors responsible for the difference so that they can take appropriate measures.
Reference
Rosner, B. (2016). Fundamentals of biostatistics. Nelson Education.