A sample of 4 widgets was taken at random from a manufacturing plant (see table below) Prior research has shown that the population has a mean of 5 and standard deviation of 1
5 |
2 |
1 |
4 |
Compute a 95 percent confidence interval for the data (2-tailed). What is its lower bound?
A sample of 4 widgets was taken at random from a manufacturing plant (see table below)...
A random sample of size n = 21, taken from a normal population with a standard deviation 04 =5, has a mean X4 = 90. A second random sample of size n2 = 37, taken from a different normal population with a standard deviation o2 = 4, has a mean X2 = 39. Find a 94% confidence interval for 11 - H2 Click here to view page 1 of the standard normal distribution table. Click here to view page 2...
A simple random sample of size n-23 is drawn from a population that is normally distributed. The sample mean is found to be x = 63 and the sample standard deviation is found to be s 18. Construct a 95% confidence interval about the population mean. The lower bound is The upper bound is (Round to two decimal places as needed.)
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A simple random sample of size n=24 is drawn from a population that is normally distributed. The sample mean is found to be x = 68 and the sample standard deviation is found to be s = 13. Construct a 95% confidence interval about the population mean. The lower bound is The upper bound is (Round to two decimal places as needed.)
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A simple random sample of size n=20 is drawn from a population that is normally distributed. The sample mean is found to be x = 59 and the sample standard deviation is found to be S = 11. Construct a 95% confidence interval about the population mean. The lower bound is . The upper bound is . (Round to two decimal places as needed.)
A sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55. Determine the 99% confidence interval for the population mean. (Round your answers to 2 decimal places.) Confidence interval for the population mean is _______ and _______ .A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal distribution with a population standard deviation...
The monthly incomes from a random sample of workers in a factory are shown below. Monthly Income, in $1,000 4.0 5.0 7.0 5.0 6.0 6.0 10.0 8.0 9.0 (Please, avoid rounding intermediate steps and round your final solutions to at least 2 decimal places) Compute the 95% confidence interval for the mean monthly incomes of the workers. Provide the lower and upper bound of the interval below and give your answer in dollars. Are there any additional assumptions needed in...
Suppose a random sample of size 17 was taken from a normally distributed population, and the sample standard deviation was calculated to be s = 5.0. a) Calculate the margin of error for a 95% confidence interval for the population mean. Round your response to at least 3 decimal places. b) Calculate the margin of error for a 90% confidence interval for the population mean. Round your response to at least 3 decimal places.
Suppose that a simple random sample is taken from a normal population having a standard deviation of 15 for the purpose of obtaining a 95% confidence interval for the mean of the population. a. If the sample size is 44, obtain the margin of error. b. Repeat part (a) for a sample size of 25
you take a random sample size of 1500 from population 1 and a random sample size of 1500 from population two. the mean of the first sample size is 76; the sample standard deviation is 20. the mean of the second sample is 62; the sample standard deviation is 18. construct the 90% confidence interval estimate of the difference between the means of the two populations representwd here and report both the upper and lower bound of the interval.
10. (12 pts) Your competitor widget maker took a random sample of 8 widgets and measured x. He got the values listed below. He is new in the business and has little knowledge of the value of population standard deviation of x. All he is going by are the 8 values he got below and an assumption that the xis distributed normally in the population: x 16.4 16.9 17.1 17.2 16.8 17.1 16.8 16.9 Compute the 95% confidence interval for...