Hypothesis Testing and Inferences
Student’s Name
Institution Affiliation
Date
Table 3 (µ = 100 seconds and σ = 30)
Child | Mean seconds of concentration in an experiment of reading | z-score | p-value |
---|---|---|---|
1 | 75 | –0.83 | 0.203 |
2 | 81 | –0.63 | 0.264 |
3 | 89 | –0.37 | 0.356 |
4 | 99 | –0.03 | 0.488 |
5 | 115 | 0.50 | 0.309 |
6 | 127 | 0.09 | 0.464 |
7 | 138 | 1.27 | 0.102 |
8 | 139 | 1.30 | 0.967 |
9 | 142 | 1.40 | 0.081 |
10 | 148 | 1.60 | 0.055 |
1st Question
Results from the standard normal table indicate that no child had come from a significantly different population (SDP) different from the one used in the null hypothesis. With consideration of the significant level of 0.05, all children had p-value greater than 0.05 (Silvey, 2017). These results would have been statistically significant if the null hypothesis was rejected under conditions that p < 0.05. In case, the p-value is greater than 0.05, and the null hypothesis would have been accepted since there is no enough evidence to reject the null hypothesis.
2nd Question
The significant level of each child becomes less at the same time with the progressive dispersion of data (Azzalini, 2017). This happens since there is a greater variation between variables.
Conclusion Statement
1st Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.203 at z < 0.83. As such, we accept the null hypothesis since p > 0.05.
2nd Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.264 at z < 0.63. As such, we accept the null hypothesis since p > 0.05.
3rd Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.356 at z < 0.37. As such, we accept the null hypothesis since p > 0.05.
4th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.488 at z < 0.03. As such, we accept the null hypothesis since p > 0.05.
5th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.309 at z < 0.50. As such, we accept the null hypothesis since p > 0.05.
6th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.464 at z < 0.09. As such, we accept the null hypothesis since p > 0.05.
7th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.102 at z < 1.27. As such, we accept the null hypothesis since p > 0.05.
8th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.0.967 at z < 1.30. As such, we accept the null hypothesis since p > 0.05.
9th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.081 at z < 1.40. As such, we accept the null hypothesis since p > 0.05.
10th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Findings from z analysis show that the p-value is 0.055 at z < 1.6. As such, we accept the null hypothesis since p > 0.05.
Table 4 (µ = 100 seconds and σ = 40)
Child | Mean seconds of concentration in an experiment of reading | z-score | p-value |
1 | 75 | –0.63 | 0.264 |
2 | 81 | –0.48 | 0.316 |
3 | 89 | –0.28 | 0.390 |
4 | 99 | –0.03 | 0.488 |
5 | 115 | 0.38 | 0.352 |
6 | 127 | 0.68 | 0.248 |
7 | 138 | 0.95 | 0.171 |
8 | 139 | 0.98 | 0.164 |
9 | 142 | 1.05 | 0.147 |
10 | 148 | 1.20 | 0.115 |
---|
1st Question
The findings from the z-table analysis provide that no child that appeared to come from a SDP other than the one used in the null hypothesis (Silvey, 2017).
2nd
Level of significant for each child tends to reduce based on their variations.
Conclusion Statement
1st Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.264 at z < 0.63. As such, we accept the null hypothesis since p > 0.05.
2nd Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.316 at z < 0.48. As such, we accept the null hypothesis since p > 0.05.
3rd Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.390 at z < 0.28. As such, we accept the null hypothesis since p > 0.05.
4th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.352 at z < 0.38. As such, we accept the null hypothesis since p > 0.05.
5th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.352 at z < 0.38. As such, we accept the null hypothesis since p > 0.05.
6th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.248 at z < 0.68. As such, we accept the null hypothesis since p > 0.05.
7th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.171 at z < 0.95. As such, we accept the null hypothesis since p > 0.05.
8th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 164 at z < 0.98. As such, we accept the null hypothesis since p > 0.05.
9th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.147 at z < 1.05. As such, we accept the null hypothesis since p > 0.05.
10th Student
The findings from the z analysis were to assess whether there was any child who appeared to come from a SDP that differed with the one used in the null hypothesis. Results from z analysis show that the p-value is 0.115 at z < 1.20. As such, we accept the null hypothesis since p > 0.05.
References
Azzalini, A. (2017). Statistical inference based on the likelihood. Routledge.
Silvey, S. D. (2017). Statistical inference. Routledge.
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