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)

saving = more money

step-by-step explanation:

if you have 100 bucks, and you want to buy a candy bar, even if you have one at home, if you choose not to buy it, you save.

~courtney

the answer is:

the correct option:

b.

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

then, using placing logarith with base 7 on both sides of the equation, we have:

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

hence, we have that the correct option:

b.

have a nice day!