Question

An irate patient complained that the cost of a doctor’s visit was too high. she randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor’s visit was $44.80. the standard deviation of the sample was $3.53. find a point estimate of the population mean. find the 95% confidence interval of the population mean. assume the variable is normally distributed.

Answer 1

The point estimate of the population mean is $44.80. The 95% confidence interval of the population mean is ($43.254, $46.346).

Step-by-step explanation:

To find the point estimate of the population mean, we will use the sample mean as the point estimate. In this case, the mean amount of money spent on each doctor's visit from the 20 patient sample is $44.80. Therefore, the point estimate of the population mean is $44.80.

To find the 95% confidence interval of the population mean, we will use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Error)

First, we need to find the critical value. Since we are assuming the variable is normally distributed, we can use the Z-table. We want a 95% confidence level, which corresponds to a Z-value of approximately 1.96.

Next, we need to calculate the standard error. The formula for the standard error is: Standard Error = Sample Standard Deviation / √(Sample Size)

Plugging in the values, we get: Standard Error = $3.53 / √(20) ≈ $0.789.

Finally, we can calculate the confidence interval: Confidence Interval = $44.80 ± (1.96 * $0.789) ≈ $44.80 ± $1.546 ≈ ($43.254, $46.346).