Confidence Interval for Population Proportion | CFA Level 1 Test Prep

99% Confidence Interval for Population Proportion

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Question

Of 900 consumers surveyed, 414 said they were very enthusiastic about a new home decor scheme. What is the 99% confidence interval for the population proportion (in percent)?

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Explanations

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A

Interval estimate can be found from p +/- z[p(1-p)/n]^0.5. Here we have n = 900, p = 414/900 = 0.46 and z = 2.58 (for 99%). Therefore 0.46 +/- 2.58*0.01661 and we get 0.42 and 0.50.

To calculate the 99% confidence interval for the population proportion, we can use the following formula:

Confidence Interval = Sample Proportion ± Margin of Error

where: Sample Proportion = Number of consumers enthusiastic / Total number of consumers surveyed Margin of Error = Z * Standard Error

The standard error is calculated using the formula:

Standard Error = √[ (Sample Proportion * (1 - Sample Proportion)) / Sample Size ]

Let's plug in the given values:

Number of consumers enthusiastic (sample size) = 414 Total number of consumers surveyed = 900

Sample Proportion = 414 / 900 ≈ 0.46 (rounded to two decimal places)

Now, we need to find the critical value (Z) corresponding to a 99% confidence level. The critical value can be found using a standard normal distribution table or calculator. For a 99% confidence level, the critical value is approximately 2.576.

Standard Error = √[ (0.46 * (1 - 0.46)) / 900 ] ≈ 0.0154 (rounded to four decimal places)

Margin of Error = 2.576 * 0.0154 ≈ 0.0397 (rounded to four decimal places)

Finally, we can calculate the confidence interval:

Confidence Interval = Sample Proportion ± Margin of Error

Confidence Interval = 0.46 ± 0.0397

Confidence Interval ≈ (0.4203, 0.4997)

To express the confidence interval in percentage terms, we can multiply the bounds by 100:

Confidence Interval ≈ (42.03%, 49.97%)

Therefore, the correct answer is:

A. 42 and 50