Question

Problem 4. (17 points) A random sample of n = 80 in-coming college freshmen showed that 28 were planning to bring a car with them to campus next fall. Suppose that we are interested in forming a confidence interval at 99.7 percent for the proportion of all in-coming college freshmen who are planning to bring a car to campus next fall. Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places. (a) The estimate is: 0.70 (b) The standard error is: 0.045 (c) The multiplier is: Note: You can earn partial credit on this problem. nroviowanoworn

Answer 1

a) estimate= Sample Proportion ,    p̂ = x/n = 28/80 = 0.35

b) Standard Error ,    SE = √[p̂(1-p̂)/n] = √(0.35*0.65/80) = 0.053

c) Level of Significance,   α =    0.003  
multiplier = z - value =   Zα/2 =    2.968   [excel formula =NORMSINV(α/2)]