Answer:
a.[144;158]$
b. The parameter estimated is the population mean of the weekly expenses on the food of American families.
Step-by-step explanation:
Hello!
The study variable for this problem is "Weekly expenses on food spent by an American family"($)
For a sample n=1014 American families the corresponding sample mean is x[bar]=$151
It is stated that for the 95%CI the margin of error is ±$7
a.
The confidence interval that estimates the population mean is constructed as the "estimator ± margin of error" for example, under a Standard Normal distribution the formula for the confidence interval is:
[x[bar]±*(δ/√n)]
For the sample taken it is:
[151±7]
The inferior limit is 144$
The superior limit is 158$
b.
The parameter estimated is the population mean of the weekly expenses on the food of American families.
The confidence level of an interval is the probability under which it is built. This probability indicates that if they build 100 confidence intervals, we expect 95 to contain the value of the population mean we are trying to estimate.
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Answer:
B) .35
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem:
Then
So the correct answer is:
B) .35