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.

(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

 from the probability table.

The margin of error (e) =

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

6.5

step-by-step explanation:

you said it was in the question

~courtney