Question

A research firm conducted a survey of 49 randomly selected Americans to determine the mean amount spent on coffee during a week. The sample mean was $20 per week. The population distribution is normal with a standard deviation of $5. What is the point estimate of the population mean? Using the 95% level of confidence, determine the confidence interval for μ.

Answer 1

Answer:

(21.4, 18.6)

Step-by-step explanation:

Given that:

Mean (μ) = $20 per week

Standard deviation (σ) = $5

number of sample (n) = 49

Confidence level = 95% = 0.95

α = 1 - 0.95 = 0.05

$\frac{\alpha }{2}=\frac{0.05}{2} = 0.025\\ z_{\frac{\alpha }{2} }=z_{0.025}= 1.96$  from the probability table.

The margin of error (e) = $z_{\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } = 1.96*\frac{5}{\sqrt{49} } =1.4$

Confidence interval = μ ± e = (μ + e, μ - e) = (20 + 1.4, 20 - 1.4) = (21.4, 18.6)