5 8 points Exercise 143 Spending on food A 2012 Gallup surve

5. (8 points) Exercise 14.3 Spending on food. 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?

Solution

a)
CI = x ± Z a/2 * (sd/ Sqrt(n)) OR x ± M.E
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=151
Sample Size(n)=1014
ME = 7
Confidence Interval = [ 151 ± 7 ] = [ 144, 158 ]
b)
Interpretations:
1) We are 95% sure that the interval [ 144, 158] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contain the rue
population mean