Question

Entered Answer Preview Result Message -2009.11, 2009.11 -2009.11, 2009.11 incorrect Your answer isn't a number (it looks like a list of numbers) 41686.9 41686.9 correct 45705.1 45705.1 correct At least one of the answers above is NOT correct. (1 point) Starting salaries of 95 college graduates who have taken a statistics course have a mean of $43,696. Suppose the distribution of this population is normal and has a standard deviation of $10,819. Using a 93% confidence level, find both of the following: (NOTE: Do not use commas nor dollar signs in your answers.) (a) The margin of error: -2009.111,2009.1 (b) The confidence interval for the mean u: 41686.889 < u < 45705.111 !!!

Answer 1

1)

a)

sample mean, xbar = 43696
sample standard deviation, σ = 10819
sample size, n = 95

Given CI level is 93%, hence α = 1 - 0.93 = 0.07
α/2 = 0.07/2 = 0.035, Zc = Z(α/2) = 1.81

ME = zc * σ/sqrt(n)
ME = 1.81 * 10819/sqrt(95)
ME = 2009.11

2)

a)

sample mean, xbar = 11.64
sample standard deviation, σ = 2.3
sample size, n = 179

Given CI level is 85%, hence α = 1 - 0.85 = 0.15
α/2 = 0.15/2 = 0.075, Zc = Z(α/2) = 1.44

ME = zc * σ/sqrt(n)
ME = 1.44 * 2.3/sqrt(179)
ME = 0.2476

b)

sample mean, xbar = 11.64
sample standard deviation, σ = 2.3
sample size, n = 179

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 2.3/sqrt(179)
ME = 0.4435

c)

Increases the margin of error