FINALSTATS1111

# FINALSTATS1111 - Exam 11 Given the following SAMPLE data...

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Exam 11. Given the following SAMPLE data calculate the indicated statistics. 10, 2, 7, 18, 9, 1, 11, 15, 15, 15 A. Mean, B. Mode, C. Median, D. Standard Deviation, E. Variance, G. Range: Use Descriptive Statistics and summary statistics F. IQR: Use the Quartile Function array: all data quart: 3 minus Use the Quartile Function array: all data quart: 1. Chevshev’s Theorem- A theorem that can be used to make statements about the proportion of data values that must be within a specified number of standard deviations of the mean. No information can be obtained falling within 1 standard deviation. At least 75% will fall within 2 standard deviations. At least 89% will fall within 3 standard deviations. The Empirical Rule- A rule that can be used to compute the percentage of data that must be within 1, 2, 3 standard deviations of the mean for data that exhibit a bell-shaped distribution. Approximately 68% of the data values will be within 1 standard deviation. 95% within 2 standard deviations 99% within 3 standard deviation. The frequency distribution- create a bin with each type of data then use the Count if function and select all the data and F4(cell reference) then select the one data from the bin. Auto sum and it should equal the sample size. Relative Frequency- Frequency divided by sample size. Percent Frequency- Relative Frequency times 100. Z-score is the mean minus the sample mean divided by the standard deviation. Classical Method- a method of assigning probabilities that assumes the experimental outcomes are equally likely. Variance - is not a measure in the same units as the original. Mean - is influenced by extreme values. Exam21. The margin of error is t * s/sqrt(n) 1. Find the following probabilities for a standard normal variable A. P( Z < -1.2 ) use NORMSDIST B. P( Z > 1.1 ) use 1-NORMSDIST C. P( 2< Z < 3 ) use NORMSDIST of 3 minus NORMSDIST 2 2. Given a normally distributed population with mean = 55 and standard deviation = 10. A. P(x < 50) use NORMDIST x= 50 mean= 55 standard deviation= 10 and cumulative will always be 1 B. P(x > 68 ) use 1- NORMDIST C. P ( 42 < x < 71 ) use NORMDIST of 71 minus NORMDIST of 42 D. 90% of the time X will be less than use NORMINV probability = .9 mean= 55 standard deviation= 10 E. 20% of the time X will be greater than use NORMINV probability= 1-.2 mean= 55 standard deviation= 10 3. A simple random sample of n=49 college professors shows the average number of hours spent using the internet per week is mean aka xbar 18 with a sample standard deviation of 10.5 hours. Estimate the 95% confidence interval for the population mean time spent using the internet per week. CI for Mean: A. The t score = use TINV probability= alpha .05(100-confidence interval) degrees of freedom = 48 (n-1) B. The margin of error is t * s/sqrt(n) Upper limit of the confidence interval= mean(xbar) + moe Lower limit of the confidence interval= mean - moe 4. A 400 students showed that 80 used on-line banking services. Estimate the 90% confidence interval for the population proportion of students

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## This note was uploaded on 03/05/2012 for the course BUSA 3131 taught by Professor Watson during the Spring '11 term at Georgia Southern University .

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FINALSTATS1111 - Exam 11 Given the following SAMPLE data...

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