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# FSU QMB 3200 - Confidence Intervals

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Confidence Intervals Point Estimator a single value that describes the population of interest x cid 0 this is easy to calculate but not always accurate interval estimates provide additional information about variability For confidence intervals we always want to assume that the sample distribution is normal so either n cid 0 30 so using the central limit theorem we assume that the sample is normal or n 30 we assume that the population is normal When cid 0 is known Standard Error of the Mean x n Standard Error of the Mean with a Finite Population x n nN 1 N To Calculate the Confidence Interval UCL x x z 2 z 2 x x x x LCL The UCL and LCL describe the range in which we have some degree of confidence that the population mean lies z cid 0 2 a Z value on the Z table cid 0 significance level 1 cid 0 level of confidence cid 0 2 amount of probability in each tail of the distribution Ex Say the claim is that the average age of a Scion driver is 26 We take a sample of 18 drivers and find that the average age is 27 5 For this we will assume that the population cid 0 is 2 3 years To construct a 95 confidence interval 1 cid 0 95 so cid 0 05 z cid 0 2 025 Our confidence level 95 our one tail of 025 975 Now look up 975 on the Z table for a value of cid 0 1 96 2 3 cid 0 18 542 Now we use the UCL LCL formula 27 5 cid 0 1 96 542 28 56 26 44 This result does not support the claim that the average age of a Scion driver is 26 because our confidence interval does not contain 26 The margin of error falls in the middle of the confidence interval ME Notice that the ME is just the second part of the formula to calculate the confidence interval z 2 x x Ex Say that the average selling price of a Blackberry by a random sample of 35 customers is 311 The population cid 0 is 35 To construct a 90 confidence interval 1 cid 0 90 so cid 0 10 z cid 0 2 05 Our confidence level 90 our one tail of 05 95 Now look up 95 on the Z table for a value of cid 0 1 645 35 cid 0 35 5 92 Now we use the UCL LCL formula 311 cid 0 1 645 5 92 311 cid 0 9 74 UCL LCL 320 74 301 26 9 74 is the ME When cid 0 is unknown we substitute S for cid 0 x s n Approximate Standard Error of the Mean Approximate Standard Error of the Mean with a Finite Population s n x nN 1 N For these problems we will use a t distribution rather than a z distribution Degrees of Freedom DF n 1 DF will tell us how we re going to use the t table To Calculate the Confidence Interval UCL x x t 2 t 2 LCL x x x x Ex Say we take a random sample of 21 baseball games The average attendance is 5038 The sample standard deviation S is 1755 To construct a 90 confidence interval to estimate the average attendance cid 0 1 90 10 DF 21 1 20 1755 cid 0 21 382 97 t cid 0 2 1 725 UCL LCL 5038 cid 0 1 725 382 97 5692 62 4377 38 Calculating CI Proportions To construct a CI since cid 0 is unknown we approximate the Standard Error of the Proportion Standard Error of the Proportion with a Finite Population p p p p p 1 n p 1 n nN 1 N To Calculate the Confidence Interval zp UCL p p 2 LCL p zp 2 p Ex The IRS reported that 62 of individual tax returns were filed electronically in 2008 A random sample of 225 tax returns were selected in 2010 From the sample 163 were filed electronically Construct a 95 CI p 163 225 7244 a 05 z 025 1 025 975 look up in the z table 1 96 cid 0 7244 x 2756 225 0298 UCL LCL 7244 cid 0 1 96 0298 7828 6656 Determining Sample Size When cid 0 is Known An increase in sample size results in a decreased margin of error n 2 2 z 2 ME x 2 Determine the sample size needed for a 95 CI when cid 0 80 and ME 10 za 2 z 05 2 z 025 1 96 1 96 102 245 86 you would round up to 246 2 80 2 Determining Sample Size When cid 0 is Unknown p 1 ME 2 n p z 2 2 p If you don t have p substitute with S the sample standard deviation What sample size is needed for a 95 CI with a MOE of 4 Say p 30 z 025 1 96 n 1 96 042 08067 0016 2 30 7 504 19 rounded up to 505 Using the Finite Population Correction Factor Remember that we use the finite population correction factor when we meet 2 criteria nN 1 N 1 We are sampling without replacement 2 n N 05 There are 310 registered republicans in our population We take a sample of 27 and find an average age of 38 7 The sample standard deviation is 6 8 Construct a 95 CI Since cid 0 is unknown we use the t table df n 1 27 1 26 t 025 2 056 6 8 cid 0 26 cid 0 310 27 309 1 3087 957 1 252 UCL LCL 38 7 cid 0 2 056 1 25 41 274 36 126 Hypothesis Testing A hypothesis is a claim being made about a population parameter or proportion Null Hypothesis H0 the status quo states a belief assumed to be true unless evidence disproves it Alternative Hypothesis H1 the opposite of the null hypothesis this is true if the null is false Ho is always cid 0 cid 0 H1 is always cid 0 The odds are always with the null hypothesis The purpose is to reach a conclusion about the population Two Tail Hypothesis Test When the alternative hypothesis contains cid 0 H0 cid 0 1 8 H1 cid 0 cid 0 1 8 Unless the sample mean is much higher or much lower than 1 8 we accept the null hypothesis One Tail Hypothesis Test Upper Tail Test Lower Tail Test cid 0 level of significance size of the rejection region Test Statistic zx x cid 0 H0 cid 0 cid 0 n look this up on the z table and find cid 0 1 96 Say cid 0 05 cid 0 2 025 this is what is in each tail 95 025 975 cid 0 1 96 are your two boundary points if zx 2 0 reject H0 if zx 2 75 reject H0 if zx 1 25 do not reject H0 Say the claim is that …

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