A school psychologist wanted to examine the effects of an incentive system on the number of in-class disruptions by four boys in 5th grade. During one week, all boys were exposed to the normal classroom environment. During week two, the boys were given an incentive for every 10 minutes that they did not disrupt the class. The mean difference in the number of disruptions between weeks one and two was 5 (SD=2 disruptions). Conduct the appropriate hypothesis test (alpha=.05) and determine if the incentive system had any effects on the boys disruptions. Write a brief paragraph to describe results using APA style.Only serious answers please, I really need help. Could you also leave me an email if I have other questions please? This would mean a lot to me!! Thank so much!
This is a Two Tailed test ... in other words, you are trying to determine if you have statistical evidence to suggest that the average change was something other than zero disruptions different between weeks 1 and 2 ... just because you saw a change of 5, doesn't mean anything ... it could just be a random fluke !
Ho (Null Hypothesis): Sample Aveage change in distruptions was zero -- no change, or X-Bar = 0
Ha (alternative): Sample Average change was not 0, or X-Bar <> 0
At the 0.05 (5%) level of significance (in total), you would have 0.025 probability in the upper end of the distribution and another 0.025 probability in the other end ... that's a total of 0.05 or 5%
Using your Z-Score table, you find the Z-Score (the number of Standard Deviations in a Standardized Normal Distribution with mean of 0 and Standard Deviation of 1) associated with 0.025 or 2.5% ... this is:
-1.959961082 Standard Deviations from the mean for the low tail and 1.959961082 standard deviations from the mean for the high tail.
So, the number of standard deviations from a presumed mean of zero (under the null hypothesis) for the sample at hand is:
( 5 - 0 ) / 2 = 2.5
That is to say, that if the Null Hypothesis of NO CHANGE in behavior were true, a legitimate value of 5 drawn purely randomly could still happen, but that would be 2.5 Standard Deviations away.
But, your upper and lower limits were only +/- 1.959961082 !!!
So 95% of the time, you would expect to see something between +/- 1.959961082 standard deviations from zero (within this range) ... but this time, you saw a value associated with a whopping 2.5 standard deviations from the presumed mean of zero ... (we'll call 1.959961082 = 1.96 here)
So you would reject Ho.
Conclusion: Based upon the sample taken, believed to be truly random and under the assumptions of normality, we have statistical evidence which would tend to suggest that the observed sample difference of 5 is statistically significant at the 95% confidence level, thereby, leading us to conclude, based upon the evidence observed, that the incentive system may have played a roll in the positive improvement of the respondents.