A. random sample of 50 four year olds attending daycare centers provided a yearly tuition average of 3987$ and the population standard deviation of $630. Find the 90% confidence interval of the true mean. and B. If a daycare center were starting up and wanted to keep tuition low, what would be a reasonable amount to charge?

To find the 90% confidence interval of the true mean, we will use the formula:

Confidence Interval = Sample Mean ± Z * (Standard Deviation / √(Sample Size))

In this case, the sample mean is $3987, the population standard deviation is $630, and the sample size is 50.

A. Calculating the Confidence Interval:
Step 1: Find the Z-value for a 90% confidence level (two-tailed test). The Z-value can be obtained from a Z-table or using statistical software. For a 90% confidence interval, the Z-value is approximately 1.645.

Step 2: Plug the values into the formula:
Confidence Interval = $3987 ± 1.645 * ($630 / √50)

Step 3: Calculate the standard error of the mean (SE):
SE = Standard Deviation / √Sample Size
SE = $630 / √50

Step 4: Apply the formula:
Confidence Interval = $3987 ± 1.645 * ($630 / √50)

Calculating the values:
Confidence Interval = $3987 ± 1.645 * ( $630 / √50)
Confidence Interval = $3987 ± 1.645 * $89.02
Confidence Interval ≈ $3987 ± $146.61

Therefore, the 90% confidence interval of the true mean is approximately $3839.39 to $4134.61.

B. To determine a reasonable amount a daycare center could charge if they wanted to keep tuition low, it would depend on various factors such as operating costs, desired profit margins, and local market conditions. The confidence interval obtained in part A provides a range within which we can be 90% confident that the true mean falls. However, it does not directly suggest a specific tuition amount. This decision requires considering factors beyond the statistical analysis, such as the daycare center's financial goals and competition in the area.