Question

Provide an appropriate answer for each of the mean confidence interval problems.

1) Construct a 94% confidence interval for the population mean, μ. Assume the population has a normal distribution. A sample of 40 part-time workers had mean annual earnings of $3120 with a standard deviation of $677. Round to the nearest dollar. (Show work)

2) In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers at a local bank take per year. Based on earlier studies it is known that the standard deviation is 12.3 days per year. How large a sample must be selected if the company wants to be 95% confident that their estimate is within 3 days of the true mean? Provide an appropriate answer for each of the proportion confidence interval problems.

3) A survey of 500 non-fatal accidents showed that 122 involved uninsured drivers. Construct a 96% confidence interval for the proportion of fatal accidents that involved uninsured drivers. (Show work)

Answer 1

1)
sample mean, xbar = 3120
sample standard deviation, s = 677
sample size, n = 40
degrees of freedom, df = n - 1 = 39

Given CI level is 94%, hence α = 1 - 0.94 = 0.06
α/2 = 0.06/2 = 0.03, tc = t(α/2, df) = 1.937

ME = tc * s/sqrt(n)
ME = 1.937 * 677/sqrt(40)
ME = 207.3425

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (3120 - 1.937 * 677/sqrt(40) , 3120 + 1.937 * 677/sqrt(40))
CI = (2913 , 3327)

2)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 3, σ = 12.3

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 12.3/3)^2
n = 64.58

Therefore, the sample size needed to satisfy the condition n >= 64.58 and it must be an integer number, we conclude that the minimum required sample size is n = 65
Ans : Sample size, n = 65 or 64

3)
sample proportion, = 0.244
sample size, n = 500
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.244 * (1 - 0.244)/500) = 0.0192

Given CI level is 96%, hence α = 1 - 0.96 = 0.04
α/2 = 0.04/2 = 0.02, Zc = Z(α/2) = 2.05

Margin of Error, ME = zc * SE
ME = 2.05 * 0.0192
ME = 0.04

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.244 - 2.05 * 0.0192 , 0.244 + 2.05 * 0.0192)
CI = (0.2046 , 0.2834)