Question

2. An economist reports that 576 out of a sample of 1,200 middle-income American households participate in the stock market. A confidence interval of [0.468, 0.492] was calculated. What confidence level was used in this calculation

Answer 1

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 $\hat p$ = x/n

The sample proportion $\hat p$ = 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:

$M.E =Z_{critical} \times \sqrt{\dfrac{\hat p (1- \hat p)}{n}}$

If the calculated confidence interval was [0.468, 0.492]

Then,

$\hat p$  - 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;

$0. 012 =Z_{critical} \times \sqrt{\dfrac{0.48 (1- 0.48)}{1200}}$

$0. 012 =Z_{critical} \times \sqrt{\dfrac{0.48 (0.52)}{1200}}$

$0. 012 =Z_{critical} \times \sqrt{\dfrac{0.2496}{1200}}$

$0. 012 =Z_{critical} \times \sqrt{2.08\times10^{-4}}$

$0. 012 =Z_{critical} \times 0.01442$

$\dfrac{0. 012}{0.01442} =Z_{critical}$

$Z_{critical} =0.8322$

From the standard normal tables,

the p - value at $Z_{critical} =0.8322$ =  0.7973

Since the test is two tailed

$1 - \alpha/2= 0.7973$

$\alpha/2= 1-0.7973$

$\alpha/2= 0.2027$

$\alpha= 0.2027 \times 2$

$\alpha= 0.4054$

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%