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A study showed that the number of local jobs for top corporations in a state follows a normal distribution. The standard deviation number of jobs for this population is 1700. Use the given random sample of 40 corporations to find the 95% confidence interval for the mean number of local jobs for top corporations in this state, Round your answers to two decimat places and use ascending order: Jobs 100 700 800 900 1000 1100 1200 1300 1900

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User ColinD
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Final answer:

The 95% confidence interval for the mean number of local jobs for top corporations in the state is between -18.05 and 3200.27.

Step-by-step explanation:

The 95% confidence interval for the mean number of local jobs for top corporations in the state can be calculated using the sample mean and standard deviation. First, we need to calculate the sample mean, which is the sum of all the jobs divided by the sample size. In this case, the sample mean is (100 + 700 + 800 + 900 + 1000 + 1100 + 1200 + 1300 + 1900) / 9 = 1091.11.

Next, we calculate the standard deviation of the sample mean, which is the population standard deviation divided by the square root of the sample size. In this case, the standard deviation of the sample mean is 1700 / sqrt(9) = 566.67. Finally, we can calculate the confidence interval by subtracting and adding the margin of error to the sample mean.

The margin of error can be calculated by multiplying the standard deviation of the sample mean by the appropriate critical value from the standard normal distribution. For a 95% confidence level, the critical value is approximately 1.96. Therefore, the margin of error is 1.96 * 566.67 = 1110.16. The confidence interval is then calculated as 1091.11 +/- 1110.16, which gives us the range of (−18.05, 3200.27).

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User Kmb
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