Question

A random sample of 14 evenings (6 to 9 P.M.) at the O’Sullivan household showed the family received an average of = 5.2 solicitation phone calls each evening. The sample standard deviation was s = 1.9. Find a 95% confidence interval for the population mean number of solicitation calls this family receives each night.

a). Calculate the Margin of Error for this scenario. [6]

b). Calculate the 95% confidence interval. [8]

c). What are the conditions necessary for this process.

Answer 1

a)
sample mean, xbar = 5.2
sample standard deviation, s = 1.9
sample size, n = 14
degrees of freedom, df = n - 1 = 13

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.16

ME = tc * s/sqrt(n)
ME = 2.16 * 1.9/sqrt(14)
ME = 1.097

b)
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (5.2 - 2.16 * 1.9/sqrt(14) , 5.2 + 2.16 * 1.9/sqrt(14))
CI = (4.1 , 6.3)

c)
Random sample is taken from normally distributed population