Statistics - violence on television?

Mitsuko Manne: Alot of people may agree that it does promote violence and in a way it may cause to the contributing factor that most people mimic or gain ideas from what is viewed on tv. In reality most producers base movies on actual events or current event affairs going on in the world. So does TV or society promote violence?? We live in a world where violence is practiced, and common in most countrys is accepted as a way of life. In the USA violence is accepted as a crime, which can be answered through law and order....Show more

Lynn Melbourne: (a). Let p1 be the population(true) proportion of Canadians who believe that there is too much violence on television and let p2 be the population(true) proportion of Americans who believe the same crap thing-(i.e., that is, there is too much violence on television!). We wish to test Ho: p1-p2 = Do=0 (i.e., p1-p2 = 0 or p1=p2). versus Ha: p1-p2≠(D0=0). (i.e., p1-p2≠0 or p1≠p2).Our sample proportion of Canadians who ! believe that there is too much violence on television= ^p1= (625/1000)= 0.625Our sample proportion of Americans who believe that there is too much violence on television = ^p2= (780/1500)= 0.52Our observed test statistic= z= [(^p1-^p2) - D0]/√{[^p1*(1-^p1)/(n1-1)]+[^p2*(1-^p2)/(n2-1)]}= [(0.625-0.52)-0]/√{[0.625*(1-0.625)/(1000-1)]+[^0.52*(1-0.52)/(1500-1)]}= 5.24 i.e., |z| = 5.24At alpha = 0.05 level of significance, our rejection point= z(alpha/2)= z(0.05/2)= z0.025= 1.96Our p-value= Twice the area on the right of the test statistic-|z| under the standard normal curve= 2*[P(z>0)-P(0= 2*(0.5-0.4990)= 2*0.001= 0.002Conclusion: Because our test statistic of |z|=5.24 is much greater than that of our rejection point- z(alpha/2)=z0.025=1.96, then we can reject Ho in favor of Ha at the alpha = 0.05 level of significance, based on our two samples from the survey. Also, since our two-tailed p value of 0.002 is less than 0.05 and is even less than 0.001, then we have strong evi! dence to conclude that there is a significant difference betwe! en the proportions of Canadians and Americans who believe that there is too much violence on television.(b).Since n*^p1= 1000*0.625=625 and n*^p2=1500*0.52= 780 are both at least greater than 5, then n is considered to be large and hence the sampling distribution of (^p1-^p2)- the probability distribution of all possible differences between the two sample proportions of Canadians and Americans believe that there is too much violence on TV will follow the z distribution.(c).Our confidence level = 99% 100(1-alpha)% = 99% 1-alpha = 0.99 alpha = 0.01 So, a 99% confidence interval estimate of the difference in the proportions of Canadian and American who believe that there is too much violence on television.= [(^p1-^p2) (+-) z(alpha/2)*√{[^p1*(1-^p1)/(n1-1)]+[^p2*(1-^p2)/(n2-1)]}]= [(0.652-0.52) (+-) z(0.01/2)*√{[^0.625*(1-0.625)/(1000-1)]+[^0.52*(1-0.52)/(1500-1)]}]= [0.132 (+-) z0.005*√{[^0.625*(1-0.625)/999]+[^0.52*(1-0.52)/1499]}]= [0.132 (+-) 2.575*√{[! 0.234375/999]+[0.2496/1499]}]= [0.08,0.18](d). The above interval indicates that we are 99% confident that the difference between the true proportions of Canadian and American who believe that there is too much violence on television should be any value lie between 0.08 and 0.18.Hope this helps....Show more