Please solve all of Part 4. slower rate than those students who are not as activ
ID: 3040927 • Letter: P
Question
Please solve all of Part 4.
slower rate than those students who are not as active. The administrator believes it takes longer for these students to graduate because of their additional responsibilities. A ra sample of 156 former student leaders from a list of graduates at ASU is compared to the overall graduation rate soplete tneir degrees at a ndom . The sample shows a mean of 4.56 years for the student leaders to e with a standard deviation of 1.23. The mean for all graduates is 4.32 Do student leaders take a significantly longer time to graduate? How can you tell? What do you tell the university administrator? PART IV According to a recent random poll conducted of Tempe residents asking for which public transportation options residents supported, 44% indicated support for the light rail system, 40% for expanded bus service and 16% supported hot-air balloon travel. The sample size was 400 Tempe residents. calculate a confidence interval for the proportion supporting light rail using a 95% confidence level. Remember, to use the formula for proportions, Show all your work 1, who support the light rail system? Can you tell the city manager that more Tempe residents support light rail than any other option? Why or why not? (2 points). If the sample size were to increase to 600 residents, what would happen to the confidence interval? Why? (1 point). 2.Explanation / Answer
1) P=0.44 ,Q=1-P= 1-0.44= 0.56 and n=400
Standard error of the mean = SEM = P*Q/n = SQRT(0.44*0.56/400)= 0.02482
= (1-CL)/2 = 0.025
Standard normal deviate for = Z = 1.960
95% confidence interval of light rail proportion
Lower bound = P - (Z*SEM) = 0.44-(1.96*0.02482)= 0.391
Upper bound = P + (Z*SEM) = 0.44+(1.96*0.02482)= 0.489
range who support light rail system=(0.391,0.489)
Yes, I can tell the city manager that more Tempe support light rail by using concept of hypothesis testing of significance for difference of proportions.
2) when n= 600
Standard error of the mean = SEM = P*Q/n = 0.020
= (1-CL)/2 = 0.025
Standard normal deviate for = Z = 1.960
Proportion of positive results = P = 0.440
Lower bound = P - (Z*SEM) = 0.44-(1.96*0.02)= 0.400
Upper bound = P + (Z*SEM) = 0.44+(1.96*0.02)= 0.480
Width of Confidence interval= (0.48-0.4)= 0.08
As sample size increases width of confidence interval get decreased.