Answer:
Confidence level = 59.46%
Step-by-step explanation:
Given that:
An economist reports that 576 out of a sample of 1,200 middle-income American households participate in the stock market.
sample mean = 576
sample size = 1200
The sample proportion [tex]\hat p[/tex] = x/n
The sample proportion [tex]\hat p[/tex] = 576/1200 = 0.48
A confidence interval of [0.468, 0.492] was calculated. What confidence level was used in this calculation?
The confidence interval level can be determined by using the formula:
[tex]M.E =Z_{critical} \times \sqrt{\dfrac{\hat p (1- \hat p)}{n}}[/tex]
If the calculated confidence interval was [0.468, 0.492]
Then,
[tex]\hat p[/tex] - M.E = 0.468
0.48 -M.E = 0.468
0.48 - 0.468 = M.E
0.012 = M.E
M.E = 0.012
NOW;
[tex]0. 012 =Z_{critical} \times \sqrt{\dfrac{0.48 (1- 0.48)}{1200}}[/tex]
[tex]0. 012 =Z_{critical} \times \sqrt{\dfrac{0.48 (0.52)}{1200}}[/tex]
[tex]0. 012 =Z_{critical} \times \sqrt{\dfrac{0.2496}{1200}}[/tex]
[tex]0. 012 =Z_{critical} \times \sqrt{2.08\times10^{-4}}[/tex]
[tex]0. 012 =Z_{critical} \times 0.01442[/tex]
[tex]\dfrac{0. 012}{0.01442} =Z_{critical}[/tex]
[tex]Z_{critical} =0.8322[/tex]
From the standard normal tables,
the p - value at [tex]Z_{critical} =0.8322[/tex] = 0.7973
Since the test is two tailed
[tex]1 - \alpha/2= 0.7973[/tex]
[tex]\alpha/2= 1-0.7973[/tex]
[tex]\alpha/2= 0.2027[/tex]
[tex]\alpha= 0.2027 \times 2[/tex]
[tex]\alpha= 0.4054[/tex]
the level of significance = 0.4054
Confidence level = 1 - level of significance
Confidence level = 1 - 0.4054
Confidence level = 0.5946
Confidence level = 59.46%
