Levels of Confidence and Intervals
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1. Find the level of confidence assigned to an interval estimate of the mean formed using the interval: x − 1.96⋅σx to x + 1.96⋅σ x .
2. The lengths of 225 fish caught in Lake Michigan had a mean of 15.0 inches. Assume that the population standard deviation is 2.5 inches.
- Give a point estimate for μ .
- Find the 90% confidence interval for the population mean length.
3. In an effort to compare college costs in State of Michigan, a sample of 36 junior students is randomly selected statewide from the private colleges and 36 more from the public colleges. The private college sample resulted in a mean of $27,650 and the public college sample mean was $11,360. Assume the annual college fees for private colleges have a mounded distribution and the standard deviation is $1725. Find the 95% confidence interval for the mean costs for private colleges.
4. What effect does an increase in the level of confidence have on the width of the confidence interval?
5. You are constructing a 95% confidence interval using the following information: n = 60, = 65.5, s = 2.5, and E = 0.7. What is the value of the middle of the interval?
A) 0.7
B) 2.5
C) 0.95
D) 65.5
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The solution examines levels of confidence and intervals for college costs in the State of Michigan.
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1. Find the level of confidence assigned to an interval estimate of the mean formed using the interval: x − 1.96⋅σx to x + 1.96⋅σ x .
Solution. We know that the z-score is z=1.96. So, the confidence level is 1-alpha=95%.
2. The lengths of 225 fish caught in Lake Michigan had a mean of 15.0 inches. Assume that the population standard deviation is 2.5 inches.
• Give a point estimate for μ .
We can use the sample mean as a point estimate for μ.
• Find the 90% confidence interval ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"
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