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 ±$7
a.
The estimated confidence interval for the population mean is formed as "estimator ± margin of error." In a Standard Normal distribution, the confidence interval formula is:
[x[bar]±
*(δ/√n)]
For the provided sample, we have:
[151±7]
The lower limit equates to 144$
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!
