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Lady_Fox [76]
3 years ago
15

Question!! if anyone knows!!!

Mathematics
1 answer:
arlik [135]3 years ago
3 0
Answer: 10
Your welcome
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Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v
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Answer:

\text {CI} = (60.54, \: 81.46)\\\\

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.

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=AVERAGE(number1, number2,....)

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\bar{x} = \$71  

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

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 s = \$ 22.35

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\text {confidence interval} = \bar{x} \pm MoE\\\\

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$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and t_{\alpha/2} 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

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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\\\\

So the required 95% confidence interval is

\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)\\\\

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)

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