A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal distribution with a population standard deviation of $5. A sample of 64 Americans revealed that X¯¯¯=$20.
A. What is the point estimate of the population mean? Explain what it indicates.
B. Using the 95 percent level of confidence, determine the confidence interval for μ. Explain what it indicates.

the main concept about point estimate is the fact that a single value derived from the sample is used directly to make inference about the population.

This simply means that for point estimation, sample estimator = population estimator

In this question of ours, our sample mean x-bar = 20 dollars, that implies that the population mean is also 20 dollars.

The implication of our answer is that an average american spends 20 dollars on dollars on coffee.

B) interval estimate.

the 95 percent confidence interval for population mean is given by the formulae below

μ = x + Zα/2 * σ/√n or μ = x - Zα/2 * σ/√n

where, μ = population mean, x = sample mean = 20, n= sample size= 64, σ= population standard deviation. Zα/2= z score at 5 percent level of significance for a two tailed test = 1.96 (note that level of significance = 1 - confidence level).

for μ = x + Zα/2 * σ/√n

μ = 20+ 1.96 (5/√64)

μ = 20 + 1.96 * (5/8)

μ = 20 + 1.96 (0.625)

μ = 20 + 1.225

μ = 21.225

μ = x - Zα/2 * σ/√n

μ = 20- 1.96 (5/√64)

μ = 20 - 1.96 * (5/8)

μ = 20 - 1.96 (0.625)

μ = 20 - 1.225

μ = 18.775

Thus at 95 percent level of significance, population interval is (18.775, 21.225).

The implication of this is that on the average , an American spends at least 18.775 dollars on coffee and at most 21.225 dollars.