Question

19. A random sample of the monthly wage of 32 hired farm workers in 1929 (the year of the great stock market crash) yielded mean $44.52 and standard deviation $1.17. An 80% confidence interval for the mean monthly wage of all such workers is about: (a) [44.35, 44.69] (b) [44.25, 44.79] (c) [44.04, 45.00] (d) [44.11, 44.93] (e) [44.18, 44.86]

20. An economist wishes to estimate, to within $50 and with 95% confidence, the mean amount spent annually by American households on clothing. Assuming the standard deviation of the amount spent is $175, the minimum sample size needed is about: (a) 33 (b) 62 (c) 39 (d) 84 (e) 48

Answer 1

19)
sample mean, xbar = 44.52
sample standard deviation, s = 1.17
sample size, n = 32
degrees of freedom, df = n - 1 = 31

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, tc = t(α/2, df) = 1.309

ME = tc * s/sqrt(n)
ME = 1.309 * 1.17/sqrt(32)
ME = 0

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (44.52 - 1.309 * 1.17/sqrt(32) , 44.52 + 1.309 * 1.17/sqrt(32))
CI = (44.25 , 44.79)

20)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 50, σ = 175

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 * 175/50)^2
n = 47.06

Therefore, the sample size needed to satisfy the condition n >= 47.06 and it must be an integer number, we conclude that the minimum required sample size is n = 48
Ans : Sample size, n = 48