...Taylor (2013) states, “We hope to obtain a small enough p-value that we are justified in rejecting the null hypothesis” (para. 3). A alternative hypothesis is also known as an experimental hypothesis. It shows that there is a observed effect for an experiment. It is what one tries to show in an indirect way by using the hypothesis testing. If the null hypothesis is rejected then one will see that the alternative hypothesis is effective. Now if the finding are not null then one will not accept the alternative of the hypothesis. Reference Aron, A., Aron, E. N., & Coups, E. J. (2009). Statistics for psychology [University of Phoenix Custom Edition eBook]. Upper Saddle River, NJ: Pearson/Prentice Hall. Retrieved from University of Phoenix, Psy315 website. Taylor, C. (2013). About.com Statistics. Retrieved from...
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...Xochitl Jacques - Smith Hypothesis Testing/ Psychological Reasons for Depression PSY315 Statistical Reasoning Amber Lupo July 6, 2015 Psychological Reasons for Depression The number of people diagnosed with depression will be different because of either biological or psychological conditions. This hypothesis testing will include a two-tailed test with the alternative hypothesis testing as there will be a difference in either biological or psychological causes, and the null would be both cause depression equally with no difference. Our hypothesis testing will be from a survey of people with biological conditions diagnosed with depression. Our method of research is through survey studies because this will provide data that cannot be observed directly and does not allow conclusions. Biological Depression Depression is quite common and attributed as a mental disorder. Biological reasons for depression are varied leaving much research to be done to discover the true cause though there has been much headway in the knowledge of brain function. Through this research, more causes and therefore more cures have been uncovered, making it a more manageable...
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...PSY315 Week3 PRACTICE PROBLEMS CHAPTER 2 11. For the following scores, find the (a) mean, (b) median,(c) sum of squared deviations,(d) variance and (e) standard deviation: 2, 2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0 added up =42 divided by 21=(2 is the mean.) 0 0 0 0 0 1 1 1 1 1 2 2 2 3 3 4 4 4 4 4 5= 2 is the median When subtracted by mean the numbers are; 0 0 -2 3 -1 2 -1 1 -2 -2 -1 2 2 2 -2 -1 2 1 2 0 -1 -2 When they are squared they are 0 0 -4 9 -1 4 -1 1 -4 -4 -1 4 4 4 -4 -1 2 1 4 0 -1 -4 added up they = 8 sum of squared deviations 8 divided by 21= 2.6 variances (2.6 (2.6) = 6.76 standard deviation 12. For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d) variance, and (e) standard deviation: 1,112; 1,245; 1,361; 1,372; 1,472 added up =6,562 divided by 5= 1312.4 the mean. 1,361 is the median When subtracted by mean the numbers are 200.4 67.4 -48.6 -59.6 -159.6 When they are squared the number are 14.1 8.2 6.9 7.7 12.6 added up they equal 49.5 sum of squared deviations) 5 divided by 49.5 = 9.9 is the variance.(9.9)(9.9)=98.01 standard deviation 13. For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d) variance, and (e) standard deviation: 3.0, 3.4, 2.6, 3.3, 3.5, 3.2 added up = 19 divided by 6 =3.1mean. 2.6 3.0 3.2 3.3 3.4 3.5 (3.2 & 3.3 added =6.5 divided by 2 =3.25 median when subtracted from mean the numbers are 16 15.6 16.4 15.7 15.5 15.8 when...
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