Engineering, 03.05.2020 14:20 coolman12062

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.

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Answer from: blake2001

A) The size of the sample they should take to estimate p with 2% margin of error and 90% confidence is; n = 1691

B) The 95% confidence interval for p is;

We are 95% confident that the true proportion of customers that are happy with their purchase is between 0.3696 and 0.4304.

A) We know that formula for margin of error is;

E = z√(p^(1 - p^)/n)

We are given;

E = 2% = 0.02

Confidence level = 90%

Now, we are not given the proportion and so we will adopt nominal p^ = 0.05

critical value at CL of 90% is 1.645.

Thus, making n the subject we have;

n = z²(p^(1 - p^))/E²

n = 1.645²(0.5 * 0.5)/0.02²

n = 1691.266

n ≈ 1691

B) n = 1000

p^ = 400/1000 = 0.4

Confidence level = 95%

Formula for confidence interval;

CI = p^ ± z√(p^(1 - p^)/n)

critical value at CL 95% is 1.96

Thus;

CI = 0.4 ± 1.96√(0.4(1 - 0.4)/1000)

CI = 0.4 ± 0.0304

CI = (0.3696, 0.4304)

We are 95% confident that the true proportion of customers that are happy with their purchase is between 0.3696 and 0.4304.

Read more about confidence interval at; link

Answer from: ambermcghee

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.

Answer from: Quest

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Answer from: Quest

Do you need to draw this out?

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