Statistical Reasoning in Psychology1. A prison psychologist recorded the number of rule infractions for 15 prisoninmates over a six-month period to be 5, 4, 2, 4, 3, 5, 2, 0, 4, 4, 5, 5, 3, 4, and 3.a.Make a frequency table.Tallynumber02345frequency///////////////12354This is not correctb. Make a histogram based on the frequency table. b.Describe in words the shape of the histogram.The histogram is uniformly distributed with mode being 4 and mean is 3.47. That is notcorrect2. Identify and solve this problem by hand.The head of public safety notices that the average driving speed at a particularintersection averages μ = 35 mph with a standard deviation of σ = 7.5 mph. After aschool speed limit sign of 20 mph is placed at the intersection, the first 40 cars travelpast at an average speed of 32 mph. Using the .01 significance level, was there asignificant change in driving speed? Answer: Null hypothesis:H0: μ = 35Alternative hypothesis: μ not equal to 35Test statistic: sqrt(n)*(xbar- μ)/σwhich under the null hypothesis follows a standard normal distribution.Critical region: Reject H0 iff the absolute value of the test statistic is greater than2.57583 ( point on a standard normal distribution above which 0.5% of thedistribution lies)Calculation: n=40σ = 7.5xbar = 32Absolute value of the test statistic = 2.529822 (This is basically correct!)Conclusion: Null hypothesis is accepted at 1% level of significance.a.Sketch the distributions involved.2.53(critical region)µ=35Where is the sketch?b.Figure the confidence limits for the 99% confidence interval.Limits = 32 2.58=323.059Upper limit = 35.06Lower limit = 28.94 (This is correct!)3. A social psychologist gave a questionnaire about concern for farm workers toseven participants before and after they attended a film about union organization of farm workers. The results are shown below with high scores meaning high concern. Using the .05significance level, do these results support the hypothesis that the film affected concern forthe lives of farm workers?Scores on the Concern MeasureParticipant Before AfterA20B74C1011D1315E85F98Ga.171114Use the five steps of hypothesis testing. (This is not CorrectH0:H1:Mean before=10.714Mean after =11Standard deviation before=3.4017Standard deviation after=5.774Z==0.0825Z-tabulated for two sided=1.96 Since 1.96>0.0825 we accept null hypothesis and conclude that film does not affectconcern for the life of farm workersb.Figure the effect size and find the approximate power of this study.(This is not correct)Power =1-BB=p (accept null hypothesis or alternative hypothesis is false.)) or10.76.7916 thus effect is 6.7916Thus B=p(3.9<<17.5).From the hypothesis testing results we can conclude that there was no effect ofthe film on the concerns for the lives of the farm workers. This means that therewas no significant difference between the two. The mean before was 10.71 whilethe mean afterwards was 11. The effect size was not significant.t(6) = 2.45, p< .054. A team of cognitive psychologists studying the effects of sleep deprivation onshort-term memory decay had eight participants stay in a sleep lab for two days. Fourparticipants were randomly assigned to a condition in which they were not permitted tosleep during that period, while the other four participants were allowed to sleep when theywanted to. At the end of the two days, the participants completed a short-term memorytask that yielded the results in the table that follows. Using the .05 significance level, didsleep deprivation reduce short-term memory?Mean Number of Letters Remembered Sleep Deprived Normal Sleep798871197a.Create the appropriate graph for this problem. Where is the graph?)b.Use the five steps of hypothesis testing.This is basically correct – but Do not reject the null hypothesis.Experiment is inconclusive.Ho: sleep deprived did not reduce short term memory.H1: sleep deprived reduces short term memory.Sleep deprived mean (=7.75Normal sleep mean (=8.75Standard deviation for sleep derivative=0.9574Standard deviation for normal sleep=1.7078H0:H1:Z==Z==0.7504Z-tabulated=1.96Since 1.96 >0.754 we accept the null hypothesis and conclude that sleep deprived did notreduce short term memory.c.Figure the effect size. (This is partially correct) Mean (sleep deprived) = 7.75Mean (sleep normal) = 8.75Effect size = = 0.722d. From the hypothesis testing results, we conclude that there was no effect of sleepdeprivation on short memory. There was no significant difference on the effectsof the two, both the normal sleep and the ones who were deprived of sleep. Themean number of letters remembered for those who were deprived of sleep was7.75 while as the mean for those with normal sleep was 8.75.Total Score 15/100Points were deducted for not providing graphs and for getting problems incorrect.