Question

Health Insurers are beginning to offer telemedicine services online that replace the common office visit. Wellpoint provides a video service that allows subscribers to connect with a physician online and receive prescribed treatments (Bloomberg Businessweek, March 4-9, 2014). Wellpoint claims that users of its LiveHealth Online service saved a significant amount of money on a typical visit. The data shown below ($), for a sample of 20 online doctor visits, are consistent with the savings per visit reported by Wellpoint.

92	34	40
105	83	55
56	49	40
76	48	96
93	74	73
78	93	100
53	82

Assuming the population is roughly symmetric, construct a 95% confidence interval for the mean savings for a televisit to the doctor as opposed to an office visit.

Answer 1

Concepts and reason

The concept of confidence interval is used here to solve the question.

Confidence interval are the interval estimates of unknown population parameter. It gives the probability that population parameter will fall in this interval.

Fundamentals

The $100\left( {1 - \alpha } \right)\%$ confidence interval for the mean can be obtained using the formula:

$CI = \bar x \pm \left( {{t_{n - 1,\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}} \right)$

Here $\bar x$ is the sample mean, $s$ is the sample standard deviation, and ${t_{n - 1,\frac{\alpha }{2}}}$ where $\left( {n - 1} \right)$ is the degrees of freedom and $\alpha$ is the level of significance.

Excel or Minitab software can be used to calculate the confidence interval.

To obtain the mean and standard deviation of the data set, follow the below mentioned steps in Excel.

Step 1: Enter the data in the Excel spreadsheet. The screenshot is shown below.

Step 2: The screenshot of the formula that is used to obtain the average is shown below.

Step 3: The screenshot of the formula that is used to obtain the standard deviation is shown below.

The mean $\left( {\bar x} \right)$ and standard deviation (s) of the provided sample are 71 and 22.35 respectively.

The degree of freedom is calculated as:

$\begin{array}{c}\\\left( {n - 1} \right) = 20 - 1\\\\ = 19\\\end{array}$

The level of significance is calculated as:

$\begin{array}{c}\\\alpha = 1 - 0.95\\\\ = 0.05\\\end{array}$

The value of ${t_{n - 1,\frac{\alpha }{2}}}$ is,

$\begin{array}{c}\\{t_{n - 1,\frac{\alpha }{2}}} = {t_{20 - 1,\frac{{0.5}}{2}}}\\\\ = {t_{19,0.025}}\\\\ = 2.093\\\end{array}$

The value is obtained from the standard t distribution table.

The confidence interval for mean can be calculated as:

$\begin{array}{c}\\CI = \bar x \pm \left( {{t_{n - 1,\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}} \right)\\\\ = \left( {71 \pm 2.093\left( {\frac{{22.35}}{{\sqrt {20} }}} \right)} \right)\\\\ = \left( {71 \pm 10.46} \right)\\\\ = \left( {60.540,81.460} \right)\\\end{array}$

Ans:

The confidence interval for the mean savings for a televisit to the doctor, as opposed to an office visit, is $\left( {60.540,81.460} \right)$ .