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 an mp3 player were happy with their purchase. a. how large a sample n should they take to estimate p with 2% margin of error, proportion 0.5 and 90% confidence?

Answer 1

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

In statistics, the margin of error is just the degree of a significant error in the outcomes of random sample surveys.

The formula of margin error is, E = z√((p-vector)(1 - (p-vector)) ÷ n)

E = 2% = 0.02

Confidence level = 90%

Now, the proportion is not given so adopt nominal (p-vector) = 0.05

The critical value at CL of 90% is 1.645.

Thus, making n the subject,

n = z²(((p-vector) × (1 - (p-vector))) ÷ E²)

n = 1.645²((0.5 × 0.5) ÷ 0.02²)

n = 1691.266

n ≈ 1691

Read more about the margin of error at

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