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. 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.

93 35 40 107 84 55 55 50 38 77 50 94 94 75 74 76 95 101 53 80

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 (to 2 decimals). 95% confidence interval: $ to $ per visit

Answer 1

To find the confidence interval we need to find the mean and sample standard deviation of the given sample:

Mean M= (93 + 35 + 40 + 107 + 84 + 55 + 55 + 50 + 38 + 77 + 50 + 94 + 94 + 75 + 74 + 76 + 95 + 101 + 53 + 80)/20
= 1426/20
Mean,M= 71.3

The sample standard deviation is calculated as:

S= √(1/20 - 1) x ((93 - 71.3)² + ( 35 - 71.3)² + ( 40 - 71.3)² + ( 107 - 71.3)² + ( 84 - 71.3)² + ( 55 - 71.3)² + ( 55 - 71.3)² + ( 50 - 71.3)² + ( 38 - 71.3)² + ( 77 - 71.3)² + ( 50 - 71.3)² + ( 94 - 71.3)² + ( 94 - 71.3)² + ( 75 - 71.3)² + ( 74 - 71.3)² + ( 76 - 71.3)² + ( 95 - 71.3)² + ( 101 - 71.3)² + ( 53 - 71.3)² + ( 80 - 71.3)²)
= √(1/19) x ((21.7)² + (-36.3)² + (-31.3)² + (35.7)² + (12.7)² + (-16.3)² + (-16.3)² + (-21.3)² + (-33.3)² + (5.7)² + (-21.3)² + (22.7)² + (22.7)² + (3.7)² + (2.7)² + (4.7)² + (23.7)² + (29.7)² + (-18.3)² + (8.7)²)
= √(0.0526) x ((470.89) + (1317.69) + (979.69) + (1274.49) + (161.29) + (265.69) + (265.69) + (453.69) + (1108.89) + (32.49) + (453.69) + (515.29) + (515.29) + (13.69) + (7.29) + (22.09) + (561.69) + (882.09) + (334.89) + (75.69))
= √(0.0526) x (9712.2)
= √(510.86172)
= 22.609

Since population sample distribution is unknown we will be using t-distribution to calculated the confidence interval as:

μ = M ± t(s_M)

where:

M = sample mean df=n-1 degree of freedom
t = t statistic determined by confidence level and degree of freedom using excel formula of t-distribution
s_M = standard error = √(s²/n)

Now,

M = 71.3 df=20-1=19
t = 2.093 ( Caluclated using excel formula for t -distribution which is =T.INV.2T(0.05,19), The t-value can also be computed using t-table shown below)
s_M = √(22.609²/20) = 5.06

μ = M ± t(s_M)
μ = 71.3 ± 2.093*5.06
μ = 71.3 ± 10.581

Now the confidence interval  for the mean savings for a televisit to the doctor as opposed to an office visit is:

{$ 60.72,$ 81.88}