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A random sample of 50 four-year-olds attending day care centers provided a yearly tuition average of $3987 and the population standard deviation of $630. Find the 90% confidence interval of the true mean

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Final answer:

The 90% confidence interval for the average yearly tuition of four-year-olds attending day care centers, with a sample mean of $3987 and a population standard deviation of $630, is approximately $3840.44 to $4133.56.

Step-by-step explanation:

To find the 90% confidence interval for the true mean yearly tuition of four-year-olds attending day care centers, we can use the formula for a confidence interval when the population standard deviation is known:

CI = µ ± (z * (σ/ √n))

Where µ is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and √n is the square root of the sample size.

Given a sample mean (µ) of $3987, a population standard deviation (σ) of $630, and a sample size (n) of 50, we first need to determine the z-score that corresponds to a 90% confidence level. A z-score for a 90% confidence level is typically approximately 1.645. We then calculate the margin of error:

Margin of Error = 1.645 * ($630 / √50)
Margin of Error ≈ 1.645 * $89.10
Margin of Error ≈ $146.56

Now we calculate the confidence interval:

Lower Limit: $3987 - $146.56 = $3840.44
Upper Limit: $3987 + $146.56 = $4133.56

Therefore, the 90% confidence interval of the true mean yearly tuition is approximately $3840.44 to $4133.56.

User Aki Nishanov
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3000   is the p-*5#%?"+(                                          
User DusteD
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