Question

Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of mobile phone were happy with their purchase.
(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?
(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find
a 95% confidence interval for p.

Answer 1

Answer:

a) Sample size = 1691

b) 95% Confidence Interval = (0.3696, 0.4304)

Explanation:

(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?

The margin of error is given by

$MoE = z \cdot \frac{\sqrt{p(1-p)} }{\sqrt{n} }$

Where z is the corresponding z-score for 90% confidence level

z = 1.645 (from z-table)

for p = 0.50 and 2% margin of error, the required sample size would be

$n = \frac{1.645^{2} \cdot 0.50(1-0.50)}{0.02^{2}} \\\\n = \frac{0.6765}{0.0004} \\\\n = 1691$

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find  a 95% confidence interval for p.

The sample proportion is

p = 400/1000

p = 0.40

z = 1.96 (from z-table)

n = 1000

The confidence interval is given by

$CI = p \pm z \cdot \sqrt{\frac{p(1-p)}{n} } \\\\CI = 0.40 \pm 1.96 \cdot \sqrt{\frac{0.40(1-0.40)}{1000} } \\\\CI = 0.40 \pm 1.96 \cdot 0.01549 \\\\CI = 0.40 \pm 0.0304 \\\\CI = 0.40 - 0.0304 \: and \: 0.40 + 0.0304\\\\CI = (0.3696 ,\: 0.4304)$

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.