MATH SOLVE

5 months ago

Q:
# 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. Wellpoint claims that users of its Live Health Online service saved a significant amount of money on a typical visit. The data shown below (s), for a sample of 20 online doctor visits, are consistent with the savings per visit reported by Wellpoint. Click on the datafile logo to reference the data. DATA file 105 56 76 93 78 83 49 48 74 93 82 55 40 95 73 100 Assuming that 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. If required, round your answers to two decimal places. $ 60.54 to $ 81.46 Icon Rey

Accepted Solution

A:

Answer:[tex]71.35-2.093\frac{22.48}{\sqrt{20}}=60.83[/tex] [tex]71.35+2.093\frac{22.48}{\sqrt{20}}=81.87[/tex] Step-by-step explanation:Information given90 34 41 106 84 53 55 48 41 75 49 97 92 73 74 80 94 102 56 83
In order to calculate the mean and the sample deviation we can use the following formulas: [tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
[tex]\bar X=71.35[/tex] represent the sample mean for the sample [tex]\mu[/tex] population mean (variable of interest)
s=22.48 represent the sample standard deviation
n=20 represent the sample size Confidence intervalThe confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value would be [tex]t_{\alpha/2}=2.093[/tex]
And replacing we got:[tex]71.35-2.093\frac{22.48}{\sqrt{20}}=60.83[/tex] [tex]71.35+2.093\frac{22.48}{\sqrt{20}}=81.87[/tex]