In a random sample of 13 people. The mean length of stay at a hospital was 6.2 days. Assume the population standard deviation is 1.7 days and the length of stay are normally distributed. Construct a 99% and a 95% confidence interval.

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Answer:

We have 99% and 95% confidence interval to be

7.1241 Or 5.276

Step-by-step explanation:

Given that n = 13

Mean = 6.2

Standard deviation = 1.7

Therefore to construct confidence interval for 99% and 95%.

Z score for 95% confidence level is 1.96

We have that:

6.2 +/- 1.96(1.7/√13)

6.2 +/- 0.9241

6.2+0.9241 or 6.2-0.9241

7.1241 Or 5.276

Answer:

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Step-by-step explanation:

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