## Health Insurers are beginning to offer telemedicine services online that replace the common office visit. Wellpoint provides a video service

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.

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5 months 2021-12-29T00:18:31+00:00 1 Answer 0 views 0

## Answers ( )

1. Answer:

$$71-2.09\frac{22.35}{\sqrt{20}}=60.554$$

$$71+2.09\frac{22.35}{\sqrt{20}}=81.446$$

So on this case the 95% confidence interval would be given by (60.554;81.446)

Step-by-step explanation:

Previous concepts

A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

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

We can calculate the mean and the deviation from these data with the following formulas:

$$\bar X= \frac{\sum_{i=1}^n x_i}{n}$$

$$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$$

$$\bar X=71$$ represent the sample mean for the sample

$$\mu$$ population mean (variable of interest)

s=22.35 represent the sample standard deviation

n=20 represent the sample size

The confidence interval for the mean is given by the following formula:

$$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$$   (1)

In order to calculate the critical value $$t_{\alpha/2}$$ we need to find first the degrees of freedom, given by:

$$df=n-1=20-1=19$$

Since the Confidence is 0.95 or 95%, the value of $$\alpha=0.05$$ and $$\alpha/2 =0.025$$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.025,19)”.And we see that $$t_{\alpha/2}=2.09$$

Now we have everything in order to replace into formula (1):

$$71-2.09\frac{22.35}{\sqrt{20}}=60.554$$

$$71+2.09\frac{22.35}{\sqrt{20}}=81.446$$

So on this case the 95% confidence interval would be given by (60.554;81.446)