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A simple random sample of 800 elements generates a sample proportion p = 0.63. (Round your answers to four decimal places.)

Provide a 90% confidence interval for the population proportion.

User Willome
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Final answer:

To calculate the 90% confidence interval for a sample proportion of 0.63 from an 800 element sample, we use the standard error and Z-score for a 90% confidence level, resulting in a confidence interval of (0.6135, 0.6465).

Step-by-step explanation:

The student has provided a sample proportion (p = 0.63) from a sample size of 800 elements and needs a 90% confidence interval. To calculate the confidence interval, one needs to use the standard error of the proportion and the critical value from the Z-distribution for the given confidence level.

Steps to calculate the 90% confidence interval:

First, calculate the standard error (SE) of the sample proportion using the formula SE = sqrt[p(1-p)/n], where p is the sample proportion and n is the sample size.

Find the Z-score that corresponds to the desired confidence level, which is the critical value. For a 90% confidence interval, the Z-score is approximately 1.645.

To find the confidence interval, compute p ± Z*SE, where p is the sample proportion, Z is the Z-score, and SE is the standard error.

The calculation will yield the 90% confidence interval for the population proportion that we can expect to contain the true population proportion 90% of the time, should the procedure be repeated multiple times under the same conditions.

Applying these steps, let's calculate the 90% confidence interval for the given sample proportion of p = 0.63:

SE = sqrt[0.63(1-0.63)/800]

SE = 0.0154 (rounded to four decimal places)

90% CI = 0.63 ± 1.645(0.0154)

90% CI = (0.6135, 0.6465) (rounded to four decimal places)

Therefore, we can say with 90% confidence that the true population proportion lies between 61.35% and 64.65%.

User Mark Van Lent
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