Question

A marketing research consultant hired by Coca-Cola is interested in determining the proportion of customers who favor Coke over other soft drinks. A random sample of 400 consumers was selected from the market under investigation and showed that 53% favored Coca-Cola over other brands.

Compute a 95% confidence interval for the true proportion of people who favor Coke.

Place your LOWER limit, rounded to 3 decimal places, in the first blank   .

Place your UPPER limit in the second blank   .

Answer 1

Solution :

Given that,

n = 400

Point estimate = sample proportion = $\hat p$ = 0.53

1 - $\hat p$ = 1 -0.53 =0.47

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96 ( Using z table )   

Margin of error = E = Z$\alpha$ / 2 * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.53*0.47) / 400)

E = 0.049

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.53- 0.049< p < 0.53+0.049

0.481< p < 0.579

The 95% confidence interval for the population proportion p is : lower limit=0.481, upper limit =0.579