Question

In: Statistics and Probability

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 2 years ago
Next > < Previous

Related Solutions

A random sample of 51 newborn babies was taken at Hospital Bella Vista. The sample mean...
A random sample of 51 newborn babies was taken at Hospital Bella Vista. The sample mean was 6.87 pounds and the standard deviation was 1.76 pounds. In previous studies, the mean newborn weight was found to be 7.5 pounds. a) Does the sample from Hospital Bella Vista indicate that the mean newborn weight is less than what was found in previous studies? Use a hypothesis test with the 7-step method and the critical value method to verify if there is...
large, private, 600-bed hospital with laboratories, operating rooms, and x-ray equipment is seeking to increase revenues...
large, private, 600-bed hospital with laboratories, operating rooms, and x-ray equipment is seeking to increase revenues The administration has decided to make a 90-bed addition on a portion of adjacent land currently used for staff parking. The administrators feel that the labs, operating rooms, and x-ray department are not being fully utilized at present and do not need to be expanded to handle additional patients. The addition of 90 beds, however, involves deciding how many beds should be allocated to...
In a different shelter, a random sample of dogs’ ages was taken and the sample mean...
In a different shelter, a random sample of dogs’ ages was taken and the sample mean was χ = 7.4 years. The standard deviation for the sample was 2.53 years. At α = 0.05 , test the claim that there is a difference between the ages of dogs at the two pet shelters.
A random sample of 12 diamonds are taken in a study of the hardness of the...
A random sample of 12 diamonds are taken in a study of the hardness of the diamond. Measurements made, yielding an average value of 48.5 with a sample standard deviation of 1.5. Fill in the blanks for the critical values for a 98% confidence interval for the variance of the hardness of the diamond.
.A simple random sample of 18 recent birth records at the local hospital was taken. In...
.A simple random sample of 18 recent birth records at the local hospital was taken. In the sample, the average birth weight was 3390 grams with a standard deviation of 179.8 grams. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution. (a) Find the 99% confidence interval for the mean birth weight of all babies born in this hospital. (b) How could we reduce the margin of error while keeping...
A simple random sample of 28 recent birth records at the local hospital was taken. In...
A simple random sample of 28 recent birth records at the local hospital was taken. In the sample, the average birth weight was 119.6 ounces and the sample standard deviation was 6.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean µ. Construct a 98% confidence interval for µ, the average birthweight of newborns. (Show work please)
From a normal population, a sample of 39 items is taken. The sample mean is 12...
From a normal population, a sample of 39 items is taken. The sample mean is 12 and the sample standard deviation is 2. Construct a 99% confidence interval for the population mean.
A random sample of 84 items is taken, producing a sample mean of 43. The population...
A random sample of 84 items is taken, producing a sample mean of 43. The population standard deviation is 5.89. Construct a 90% confidence interval to estimate the population mean.
A sample of size 10 is taken from the first population: Sample mean of 101.2 and...
A sample of size 10 is taken from the first population: Sample mean of 101.2 and sample variance of 18.1 A sample of size 14 is taken from the second population: Sample mean of 98.7 and sample variance of 9.7 1)In order to decide whether pooling is appropriate or not, performing a test at α = 0.2 level of significance : Find the rejection region. 2)In order to decide whether pooling is appropriate or not, performing a test at α...
A sample of size 10 is taken from the first population: Sample mean of 101.2 and...
A sample of size 10 is taken from the first population: Sample mean of 101.2 and sample variance of 18.1 A sample of size 14 is taken from the second population: Sample mean of 98.7 and sample variance of 9.7 1)In order to decide whether pooling is appropriate or not, performing a test at α = 0.2 level of significance : Find the rejection region. 2)In order to decide whether pooling is appropriate or not, performing a test at α...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT