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Hypothesis Testing and Z-Score Analysis in Statistical Studies

A statistical study focusing on hypothesis testing and z-score analysis to evaluate data and draw conclusions.

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    Hypothesis Testing and Z - Score Analysis in Statistical Studies Statistics 11. The five steps of hypothesis testing are : (1) Specification of the Null H ypothesis ( H 0 ) and the Alternative Hypothesis ( H a ) according to the populations (2) Determine the significance level (3) Calculation of the test statistic (4) Calculation of the p - val ue (5) Conclusion : discard or not the Null Hypothesis (1) This step is necessary to define the hypothesis and to recognize the populations that will be considered throughout the study . (2 ) This step is significant: the significant level gives a position when the Null Hypothesis will be turned down. For example we can choose 5% (or p < 0.05) (3 ) Calculation of the test statistic. For instance in the case of the Z distribution the z - score is calculated as followed : 𝑧 = π‘₯ βˆ’ πœ‡ 𝜎 (4) Calculation of the p - value, which is the probability of the test statistic. This step is used t find the rejection region. (5) The final step compares the score value found previously to the cutoff score to say if there is (or there is not) enough evidence that we can reject (or cannot reject) the Null Hypothesis. 14. For these 4 studies we reject the Null Hypothesis in the case: z - score value > z - score cutoff value. A) For the study A : the z - score is 𝑧 = 7 βˆ’ 5 1 = 2 With a 5% significance level the z - score is 𝑧 0 = + 1 . 64 then the z - score cutoff value is +1.64. Therefore 2 > 1.64 so we reject the Null Hypothesis B) For the study B : the z - score is 𝑧 = 7 βˆ’ 5 1 = 2 With a 5% significance level the z - score is 𝑧 0 = Β± 1 . 96 (two tailed test) then the z - score cutoff value is Β± 1 . 96 Therefore 2 > + 1. 96 so we reject the Null Hypothesis C ) For the study C : the z - score is 𝑧 = 7 βˆ’ 5 1 = 2 With a 1% significance level the z - score is 𝑧 0 = + 2 . 33 then the z - score cutoff value is + 2 . 33 Therefore 2 < 2.33 so we fail to reject the Null Hypothesis D ) For the study D : the z - score is 𝑧 = 7 βˆ’ 5 1 = 2 With a 1% significance level the z - score is 𝑧 0 = Β± 2 . 58 (two tailed test) then the z - score cutoff value is Β± 2 . 58 Therefore - 2.58 < 2 < 2.58 so we fail to reject the Null Hypothesis

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