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Final

# BSB123 - Data Analysis Final Exam Notes (Summary of Lectures 7 - 12)

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School
Queensland University of Technology
Department
Management and Human Resources
Course
BSB123
Professor
All
Semester
Spring

Description
Data Analysis Final Exam Notes Lecture 7: Sampling Distributions Statistical inference: When a sample is selected to draw conclusions about a population Sampling distribution: Distribution of possible values any sample statistic may take or spread around the population parameter of interest Sample error: Different samples of the same size from the sample population will yield different sample means Standard error of the mean:   X  n As n increase, the sample error decreases (the more accurate the results will be) If the Population is Normal Population is normal with a menand standard deviation,  The sampling distribution Xf  and  X  n Z Value for Sampling Distribution of the Mean Z  (X  )  n Example: A soft drink manufacturer sells one of its population flavours in a 600mL bottle. Fill of the soft drink is normally distributed with a mean fill of 600mL and a standard deviation of fill of 10mL. How likely is it we would get a mean fill from a sample of 25 bottle which has a result of 598mL or less?  X    600     10  2 X n 25  X  x  P(X  598)  P      x   P Z  598600   2    x  598  P(Z  1)  0.1587 There is a 15.87% chance a sample of 25 bottles would produce a sample mean fill of less than 598mL. Christina Meyers BSB 123 Data Analysis 1 If the Population is not Normal Sample size 30 Central Limit Theorem applies Central Limit Theorem: Regardless of the shape of individual values in the population distribution, as long as the sample size is large enough the sampling distribXtwill be approximately  normally distributed withX   and X  n Example: For a non-normal population wi  8 and   3, what is the probability that the sample mean is between 7.8 and 8.2 if a sample size n = 36 is selected The central limit theorem can be usd30s The sampling distribution xis approximately normal with a mean, of 8 and standard error, X  3  X    0.5 n 36      7.88 8.28  P(7.8  X  8.2)  P 3  Z  3     36 36   P(0.4  Z  0.4)  0.3108 Sample Distribution of Proportion – Z Distribution  is population proportion pis sample proportion p  X  no.of items inthe samplehavingthe characteristicof interest n samplesize Sample distribution is binomial It can be approximated by normal n  5 and n(1)  5 With resulting mean equal to:p  Christina Meyers BSB 123 Data Analysis 2 Standard error:  (1) p n Z Value for Sampling Distribution of Proportion p  Z  (1) n Example: If the true proportion of voters who support Proposit  0.4, what is the probability that a sample size 200 yields a sample proportion between 0.40 and 0.45?  p   0.4   (1) p n 0.4(10.4)  200  0.03464 0.400.40 0.450.40  P(0.40  p  0.45)  P  Z    0.03464 0.03464   P(0  Z 1,44)  0.4251 Christina Meyers BSB 123 Data Analysis 3 Lecture 8: Estimation Point estimate: value of a single sample statistic (“best number guess”) used to estimate an unknown population parameter Confidence interval: range of values around point estimate Interpretation e.g 95% CI – If we make 100 confidence intervals we expect 95 to contain the true population parameter (+ long term frequency link) Mean point estimate for  x General Formula: Point Estimate ± (Critical Value)*(Standard Error) Confidence Level: 1 -  in each tail 2 Confidence interval  For ( Known) – Z Distribution:  Z  n Where: x is the point estimate Z is the normal distribution critical value for a probability of /2 in each tail  is the standard error n Example: Christina Meyers BSB 123 Data Analysis 4 Common Levels of Confidence Example: A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is 0.35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.  x  Z  n  0.35   2.201.96   11   2.20 0.2068 1.9932   2.4068 We are 95% confident that the true mean resistance is between 1.9932 and 2.4068. Although the true mean may or may not be in this interval, 95% of intervals formed in this manner (in repeated samples) will contain the true mean. Determining Sample Size for the Mean z  2 Sample size n  2 e Example: If = 45, what sample size is needed to estimate the mean withi 5 with 90% confidence? 2 2 2 2 n  z   (1.645) (45)  219.19 e2 52 The required sample size is n = 220 (ALWAYS ROUND UP) S For  ( Unknown) – t Distribution: x  tn1 n Where t has df = n -1 Example: A random sample of n = 25 has x = 50 & S=8. Form a 95% confidence interval fr: df = n – 1 = 25 – 1 = 24 Christina Meyers BSB 123 Data Analysis 5 t 2,n1 t0.025,242.0639 The confidence interval is S 8 x   2,n1  50  (2.0639 ) n 25 2 2 z  Sample size n  2 e p(1 p) Proportion point estimate for : p  Z  n Where: Z is the standard normal value for the level of confidence required p is the sample proportion n is the sample size Example: A random sample of 100 people shows that 25 are left handed. Form a 95% confidence interval for the true proportion of left-handers. p(1 p) p  Z  n 25 0.25(0.75)  1.96 100 100  0.251.96(0.0433) 0.1651   0.3349 Estimating Sample Size for 2 Sample size n  Z (1) e 2 NB: If completely unknown use   0.5 as the default How large a sample would be necessary to estimate the true proportion defective in a large population within  3% with 95% confidence? (Assume a pilot sample yields p = 0.12) Z (1) (1.96) (0.12)(10.12) n  2  2  450.74 e (0.03) Therefore required sample size is n = 451. Christina Meyers BSB 123 Data Analysis 6 Lecture 9: Hypothesis Testing – One Sample Tests Hypothesis: Statement about a population parameter. It uses the sample statistic as evidence to back up the claim about the population parameter. It always conta  Steps in Hypothesis Testing 1. State the null and alternative hypothesis H0: Null hypothesis – status quo H1: Alternative hypothesis – hypothesis of change  2. Choose the level of significance,and sample size, n Level of significance- total rejection area 3. Determine the appropriate test statistic and sampling distribution Means  Known – Z Distribution Test statistic: Z  x    n Means  Unknown – t Distribution Test statistic: x   t  S n Proportions – Z Distribution p  Z  (1) n 4. Determine the critical values that divide the rejection and non-rejection regions Decision diagram technique One tail test – Upper tail – Direction Implied ‘increase’ or ‘d1 reject region H :   x 0 H 1   x Upper tail test 1f H is focused on above the mean Christina Meyers BSB 123 Data Analysis 7 One tail test – Lower tail– Direction Implied ‘increase’ or ‘decrease’1 reject region H :   x 0 H :   x 1 Is a lower-tail test i1 H is focused on below the mean Two tail test – No Direction Implied 2 reject regions H :0  x H :   x 1 p-Value technique p-Value: Probability of obtaining test statistic more extreme than observed value given 0 is true If p-Value
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