A 2012 Gallup survey of a random sample of 1014 American adults indicates that American families spend on average $151 per week

Question

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A 2012 Gallup survey of a random sample of 1014 American adults indicates that American families spend on average $151 per week

on food. The report further states that, with 95% confidence, this estimate has a margin of error of ±$7.
(a) This confidence interval is expressed in the following form: "estimate ± margin of error." What is the range of values (lower bound, upper bound) that corresponds to this confidence interval?

(b) What is the parameter captured by this confidence interval? What does it mean to say that we have "95% confidence" in this interval?

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-04-07T05:25:26+03:00

Answer:

a.[144;158]$

b. The parameter being estimated is the average weekly food expenses of American families.

Step-by-step explanation:

Hello!

The variable studied here represents "Weekly food expenses incurred by an American family"($)

From a sample of n=1014 American families, the sample mean is x[bar]=$151

It is indicated that for the 95% confidence interval, the margin of error is ±$7

a.

The estimated confidence interval for the population mean is formed as "estimator ± margin of error." In a Standard Normal distribution, the confidence interval formula is:

[x[bar]± $Z_{\alpha/2}$ *(δ/√n)]

For the provided sample, we have:

[151±7]

The lower limit equates to 144$

The upper limit amounts to 158$

b.

The parameter being contemplated is the average weekly food expenses of American families.

The confidence level for this interval reflects the likelihood that if 100 such intervals were constructed, approximately 95 would encompass the true value of the population mean we aim to estimate.

Wishing you a wonderful day!