Answer:
[tex]\text {CI} = (60.54, \: 81.46)\\\\[/tex]
Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)
Step-by-step explanation:
Let us find out the mean savings for a televisit to the doctor from the given data.
Using Excel,
=AVERAGE(number1, number2,....)
The mean is found to be
[tex]\bar{x} = \$71[/tex]
Let us find out the standard deviation of savings for a televisit to the doctor from the given data.
Using Excel,
=STDEV(number1, number2,....)
The standard deviation is found to be
[tex]s = \$ 22.35[/tex]
The confidence interval is given by
[tex]\text {confidence interval} = \bar{x} \pm MoE\\\\[/tex]
Where the margin of error is given by
[tex]$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\[/tex]
Where n is the sample of 20 online doctor visits, s is the sample standard deviation and [tex]t_{\alpha/2}[/tex] is the t-score corresponding to a 95% confidence level.
The t-score is given by is
Significance level = α = 1 - 0.95 = 0.05/2 = 0.025
Degree of freedom = n - 1 = 20 - 1 = 19
From the t-table at α = 0.025 and DoF = 19
t-score = 2.093
So, the margin of error is
[tex]MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\[/tex]
So the required 95% confidence interval is
[tex]\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\[/tex]
Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)
