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This refers to a positive whole number that indicates the lack of restrictions in calculations. The following states below:
(rows-1)(columns-1) = (4-1)(5-1)= 3= 12
Critical Value: In hypothesis testing, the critical value is a number that separates a region where the null hypothesis will not be rejected from a region where it will be rejected. It is achieved at a particular percentile range and is then determined using a chart, known as the distribution topic.” The equation below states the critical value at 5% significance level:
C.95 =21.026
* The .95 is taken from 100%-5%=95% *The 12 is from the degrees of freedom
Lastly, I found what conclusion could be drawn regarding the grades achieved on a semester basis and the time spent on a daily basis on social media sites. I used information from stating my null and alternative hypothesis to either accept/reject my null/alternative. I did this through comparing my critical value at 5% and my chi-squared . My outcome is stated below:
cal < co (df) = accept the Ho cal > co (df) = reject Ho and accept HI
= 54.33 C.95 =21.026 …show more content…
It is puzzling that students could spend as much as two hours or more and still achieve an average of an A. However, in order to conclude that all of the following surveys are 100% accurate, extensive research would have to be completed (regarding academic backgrounds, level of classes taken AP/CP, increase in number of students surveyed, etc.). One subset of data defying the trend were the surveys that showed 13 students achieved an A and yet spent more than three hours in total on average per night on social media sites. However, then one must take into account the student’s intelligence level and how much time he or she spends sleeping, studying, etc. This led to uneven data and possibly skewed results. It is evident though, that based on this chart and the final secured evidence, that this does not affect my
(rows-1)(columns-1) = (4-1)(5-1)= 3= 12
Critical Value: In hypothesis testing, the critical value is a number that separates a region where the null hypothesis will not be rejected from a region where it will be rejected. It is achieved at a particular percentile range and is then determined using a chart, known as the distribution topic.” The equation below states the critical value at 5% significance level:
C.95 =21.026
* The .95 is taken from 100%-5%=95% *The 12 is from the degrees of freedom
Lastly, I found what conclusion could be drawn regarding the grades achieved on a semester basis and the time spent on a daily basis on social media sites. I used information from stating my null and alternative hypothesis to either accept/reject my null/alternative. I did this through comparing my critical value at 5% and my chi-squared . My outcome is stated below:
cal < co (df) = accept the Ho cal > co (df) = reject Ho and accept HI
= 54.33 C.95 =21.026 …show more content…
It is puzzling that students could spend as much as two hours or more and still achieve an average of an A. However, in order to conclude that all of the following surveys are 100% accurate, extensive research would have to be completed (regarding academic backgrounds, level of classes taken AP/CP, increase in number of students surveyed, etc.). One subset of data defying the trend were the surveys that showed 13 students achieved an A and yet spent more than three hours in total on average per night on social media sites. However, then one must take into account the student’s intelligence level and how much time he or she spends sleeping, studying, etc. This led to uneven data and possibly skewed results. It is evident though, that based on this chart and the final secured evidence, that this does not affect my