Question

(18.02) A survey of licensed drivers inquired about running red lights. One question asked, "Of every ten motorists who run a red light, about how many do you think will be caught?" The mean result for 873 respondents was x¯¯¯ = 2.08 and the standard deviation was s = 1.74. For this large sample, s will be close to the population standard deviation s, so suppose we know that s = 1.74.

Give a 95% confidence interval (±0.01) for the mean opinion in the population of all licensed drivers (_____+- _____)

The distribution of responses is skewed to the right rather than Normal. Will it strongly affect the z confidence interval for this sample?

Yes

No

Answer 1

sample mean, xbar = 2.08
sample standard deviation, s = 1.74
sample size, n = 873
degrees of freedom, df = n - 1 = 872

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.963

ME = tc * s/sqrt(n)
ME = 1.963 * 1.74/sqrt(873)
ME = 0.116

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2.08 - 1.963 * 1.74/sqrt(873) , 2.08 + 1.963 * 1.74/sqrt(873))
CI = (1.964 , 2.196)

yes