Answer:
The answer is below
Explanation:
a)
Given that mean (μ) = $1500, standard deviation (σ) = $200, sample size (n) = 100
confidence (C) = 95% = 0.95
α = 1 - C = 1 - 0.95 = 0.05
α/2 = 0.05 / 2 = 0.025
The z score that corresponds with 0.475 (0.5 - 0.025) is 1.96. Therefore the margin of error (E) is:
[tex]E = z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{200}{\sqrt{100} } =39.2\\[/tex]
The confidence interval = (μ ± E) = (1500 ± 39.2) = (1500 - 39.2, 1500 + 39.2) = (1460.8, 1539.2)
The confidence interval is between $1460.8 and $1539.2.
b) Given that mean (μ) = $1500, standard deviation for 100 samples = σ /√n = $200,
confidence (C) = 95% = 0.95
[tex]E = z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*200=392\\[/tex]
The confidence interval = (μ ± E) = (1500 ± 392) = (1500 - 392, 1500 + 392) = (1108, 1892)
The confidence interval is between $1108 and $1892.
