A health insurer wants to know if having a nurse available to answer simple questions over a telephone hotline could cut costs by eliminating unnecessary doctor visits. Records show the yearly cost of doctor visits has an SD of $200. They randomly select 100 families, give them access to the hotline, and record the costs of their doctor visits. In this sample, they find the average yearly cost per family to be $850. Find a 95% confidence interval for the yearly population average cost of doctor visits per family.Answer Approximately from $830 to $870 Approximately from $820 to $860 Approximately from $810 to $890 Approximately from $790 to $910
The 95 % confidence interval, CI, for a mean is :
Sample mean – z (0.05)*s/SQRT(n) < True mean < Sample mean + z (0.05)*s/SQRT(n)
where z(.05) is the value from the standard normal distribution corresponding to a probability of 0.05 of a larger value (sign ignored), s is the sample standard deviation, and SQRT(n) is the square root of the sample size n. If the sample size is less than 30 it is better to use the t distribution than the z distribution.
For your case then,
850 – 1.96*200 / SQRT(100) < true mean < 850 + 1.96*200 / SQRT(100)
850 – 39.2 < True mean < 850 + 39.2
810.8 < True mean < 889.2
The answer is approximately from $810 to $890.If you have questions please ask. High positive rating and bonus would be greatly appreciated. Thanks for the question.
BS mathematics, MS biostatistics, 35+ yrs designing & analyzing biological experiments.