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Statistics Problem Set: Developing Confidence Intervals

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1. A car rental office checks out an average of 30 cars per day with a standard deviation of 8 cars. If a sample of 25 days of operation is selected and the sample mean is computed. What is the value for the standard error of the mean? What is the probability that the sample mean for the 25 days will be within 27 and 32 cars?

2. In an effort to estimate the mean annual amount spent per customer for groceries at a particular supermarket, data were collected for a sample of 100 households. The sample showed an average amount of $8,000. If the population standard deviation is $500. Develop an 80% then a 85% confidence interval estimate of the population mean annual amount spent. If the data were collected for a sample of 50 households rather than 100, develop an 85% confidence interval estimate of the population mean annual amount spent, assuming the population has a normal probability distribution

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The expert develops confidence intervals and problem sets.

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1. A car rental office checks out an average of 30 cars per day with a standard deviation of 8 cars. If a sample of 25 days of operation is selected and the sample mean is computed,

(a) What is the value for the standard error of the mean?

Solution. Given that x-bar = 30, sigma = 8 and n = 25. Using a formula for the standard error of the mean: sigma_x-bar = sigma/(sqrt(n)) = 8/(sqrt(25)) = 1.6.

(b) What is the probability that the sample mean for the 25 days will be within 27 and 32 cars?
Solution. Let Z = (x-bar - 30)/1.6 . Then Z~N(0, 1). ...

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  • 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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