5. For a sample of 8 operating rooms taken in a hospital study , the mean...

5. For a sample of 8 operating rooms taken in a hospital study , the mean noise level was 39.89 decibels and the standard deviation was 11.1. find the 99% confidence interval of the true mean of the noise levels in the operating rooms.Assume the variable is joe ally disturbed. Round your answers to two decimals places ______<u<_______

The speeds in miles per hour of seven randomly selected qualifiers for the Indianapolis 500 (in 2012) are listed below. Estimate the mean qualifying speed with 90% confidence. Round your answers to four decimal places.

222.891 222.929 223.422 223.684 224.037 225.172 226.40

_______ < u < ______________

7.the number of unhealthy days based on the AQi for a random sample of metropolitan area is shown. Round the sample statistics and final answers to one decimal place.

5 40 61 39 1 0 16 29 23 10

Construct a 95% confirmed interval based on the data . Assume the variable is normally distributed.

Solutions

Expert Solution

Quesiton 5

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 8- 1 ) = 3.499
39.89 ± t(0.01/2, 8 -1) * 11.1/√(8)
Lower Limit = 39.89 - t(0.01/2, 8 -1) 11.1/√(8)
Lower Limit = 26.16
Upper Limit = 39.89 + t(0.01/2, 8 -1) 11.1/√(8)
Upper Limit = 53.62
99% Confidence interval is ( 26.16 , 53.62 )

Question 6

Mean X̅ = Σ Xi / n
X̅ = 1568.535 / 7 = 224.0764
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 9.9049 / 7 -1 ) = 1.2848

The mean qualifying speed is 224.0764

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 7- 1 ) = 1.943
224.0764 ± t(0.1/2, 7 -1) * 1.2848/√(7)
Lower Limit = 224.0764 - t(0.1/2, 7 -1) 1.2848/√(7)
Lower Limit = 223.1329
Upper Limit = 224.0764 + t(0.1/2, 7 -1) 1.2848/√(7)
Upper Limit = 225.0199
90% Confidence interval is ( 223.1329 , 225.0199 )

Question 7

Mean X̅ = Σ Xi / n
X̅ = 224 / 10 = 22.4
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 3576.4 / 10 -1 ) = 19.9343

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 10- 1 ) = 2.262
22.4 ± t(0.05/2, 10 -1) * 19.9343/√(10)
Lower Limit = 22.4 - t(0.05/2, 10 -1) 19.9343/√(10)
Lower Limit = 8.1
Upper Limit = 22.4 + t(0.05/2, 10 -1) 19.9343/√(10)
Upper Limit = 36.7
95% Confidence interval is ( 8.1 , 36.7 )

orchestra answered 1 year ago

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