Question

1. Find the 90% Cl for the population mean if sample

Answer 1

Answer:

The answer is below

Step-by-step explanation:

1)

mean (μ) = 12, SD(σ) = 2.3, sample size (n) = 65

Given that the confidence level (c) = 90% = 0.9

α = 1 - c = 0.1

α/2 = 0.05

The z score of α/2 is the same as the z score of 0.45 (0.5 - 0.05) which is equal to 1.65

The margin of error (E) is given as:

$E=Z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} } =1.65*\frac{2.3}{\sqrt{65} } =0.47$

The confidence interval = μ ± E = 12 ± 0.47 = (11.53, 12.47)

2)

mean (μ) = 23, SD(σ) = 12, sample size (n) = 45

Given that the confidence level (c) = 88% = 0.88

α = 1 - c = 0.12

α/2 = 0.06

The z score of α/2 is the same as the z score of 0.44 (0.5 - 0.06) which is equal to 1.56

The margin of error (E) is given as:

$E=Z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} } =1.56*\frac{12}{\sqrt{45} } =2.8$

The confidence interval = μ ± E = 23 ± 2.8 = (22.2, 25.8)