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## Statistical Methods

by: Bailey Macejkovic

31

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27

# Statistical Methods STAT 51100

Marketplace > Purdue University > Statistics > STAT 51100 > Statistical Methods
Bailey Macejkovic
Purdue
GPA 3.63

Staff

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COURSE
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27
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KARMA
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## Popular in Statistics

This 27 page Class Notes was uploaded by Bailey Macejkovic on Saturday September 19, 2015. The Class Notes belongs to STAT 51100 at Purdue University taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/207944/stat-51100-purdue-university in Statistics at Purdue University.

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Date Created: 09/19/15
511 Statistical Methods Purdue University Dr Levine Fall 2006 Practice Problems Davore Chapters 18 Aua 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 1 Pr 22 o A very large percentage of the data values are greater than 0 which indicates that most but not all runners do slow down at the end of the race a The histogram is also positively skewed which means that some runners slow down a lot compared to the others a A typical value for this data would be in the neighborhood of 200 seconds The proportion of the runners who ran the last 5 km faster than they did the first 5 km is very small about 1 or so AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Ch2 Pr9 a We want to choose all of the 5 cordless and 5 out of 10 others thus the desired probability is 5 10 5 5 252 15 10 3003 0 Fix one group say the cordless phones We want the other two groups represented in the last 5 serviced Thus we choose 5 out of 10 others also we don t want to include the outcomes where the last 5 are all the same So we have 150 2 155 when the group is fixed AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 o In general case it becomes 3 m 2 ilt g i Z 5 0 Now we want to choose 2 of the 5 cordless 2 of the 5 cellular and 2 of the corded phones SKEW 1 1998 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 2 Pr 13 PA m A m A3 PA3 PA1 m A3 PA2 0 A3 PA1 m A2 0 A3 17 PA1 Ag u A3 1314 1 m A m 149 PA3 1 PA1 U A2 U A3 PA3 1 PA1PA2PA3 PA1 A2 PA1 m A3 PA2 m A3 PA1 m A2 m 143 PA3 75 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter2 Pr 58 0 Just using probability axioms and the definition of conditional probability we find that PA u B m C PA n C o B n 0 PAUBC 130 130 PAnCPBrC PAnBnC 130 PAC PBC PA m BlC AUG 2006 511 Statistical Memods Purdue University Dr Levine Fall 2006 Problem 71 Chapter 2 n Just mm that PM n B 133 PA n B PB PA 133 1 PA w Pm PM 1309 Ann 2066 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Problem 76 Chapter 2 o The desired condition is P at least one opens 1 055 999969 a The next one is P at least one fails to open 1 955 2262 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 2 Pr 18 a Note that this pr requires the complement concept a Let event A be that the 75 W bulb is selected first Its probability is 1 Then what we need to find is PA PA 1 PA 6 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 3 Pr 33 a Note that EXZ39PZ CZ 21 21 21 a It is a wellknown fact from the infinite series theory that AUG 2006 511 Statistical Memods Purdue University Fall 2006 Dr Levine Chapter 3 Fr 37 a Note that 6 EhX E 2 408 G a At the same time a 28 Ann 2066 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 3 Pr 53 o LetX the number of flashlights that work Let event B battery has acceptable voltage 0 Then P flashlight works P both batteries work PB2 081 0 Thus X N Bin10 81 and PX 2 9 PX 9 PX 10 190 819 Cg8110 407 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter3 Pr 58 0 Since VX np1 p the only values Of that turn the variance into zero are 1 and 0 Both correspond to the situation where nothing is random 0 Take the derivative of np1 p with respect to p to put it equal to zero to obtain n 2pm O This produces p 05 a Note that you need to verify thatp 05 is indeed the point of maximum of the function np1 p AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Chapter 3 Pr 83 o For twohour period the parameter of the distribution is At 4 2 8 Therefore 10 2 PX1 0 e 10 0099 a Fora 30 minute period At 4 2 Therefore 0 2 PX 0 8 2E 135 Clearly EX At 2 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Lecture 16 Tests about a Population Proportion Devore Section 83 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 LargeSample Tests 0 Letp denote the proportion of individuals or objects in a population who possess a specified property thus each object either possesses a desired property 8 or it doesn t F 0 Consider a simple random sample X1 7Xn If the sample size n is small relative to the population size the number of successes in the sample X has an approximately binomial distribution If n itself is also large both X and the sample proportion 13 Xn are approximately normally distributed a Largesample tests concerning p are a special case of the more general largesample procedures for an arbitrary parameter 6 We considered such a largesample test before for the mean u of an arbitrary distribution Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 Some basic properties of are 1 Estimator is unbiased E13 p 2 Second it is approximately normal and its standard deviation SD is Up p1 pn 3 Note that 013 does not include any unknown parameters This is not always the case It is enough to remember the largesample test of the mean where 0 0X 0271 which is in general unknown unless 02 is specified Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 a Let us consider first an uppertailed test It means having a null hypothesis H0 p p0 vs an alternative Ha p gt 1 0 Under the null hypothesis we have E p0 and 0p xp01 p0n therefore for large n the test statistic i3 p0 p01 p0n has approximately standard normal distribution a The rejection region is clearly z 2 Za for a test of approximately level 05 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 o The lowertailed test has a rejection region z g za 0 The twotailed test has a rejection region 2 Zag The last expression is a concise way of saying that z 2 Z042 or 2 g za2 0 These tests are applicable whenever the normal approximation of the binomial distribution is reasonable npo 2 10 n1 p0 gt 10 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Example 0 Consider example 811 from Devore The null hypothesis here is thatp 03 The alternative would be p gt 30 a Note that the rule of thumb is satisfied npo 41153 gt 10 abd n1 p0 41157 gt 10 0 Thus we use the largesample test with Z I 03 3 37n a For a significance level 05 1 we use za 128 The sample proportion i513 12764115 310 Plugging this value in Z we obtain 140 gt 128 Thus the null hypothesis is rejected Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Type II Error and sample size determination a Type II Error probability can be computed exactly as before If H0 is not true the true proportion p pl 7 p0 Under Ha p pl we have Z is still approximately normal however amp EltZgt 1901 P0n and 29 1 p n VltZgt 1901 P0n Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 o The formulas for the type II error are very similar to what we saw before for the mean test We only give the uppertailed test formula Ha p gt p0 I p0pZa P01 P0n vu pvn and the lowertailed test formula Ha p lt p0 290 25 in po1 pon 1 Q pu pvn a Sample size formulas can also be easily derived In the twotailed case the formula is approximate as before Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Example 0 Consider example 812 from Devore The null hypothesis is H0 19 09 vs Ha p lt 09 How likely is it thata test of level 01 based on n 225 packages detect a departure of 10 from the null value a Forp 08 we have MB1 8 233BXJVm5 r8r2225 0228 0 Thus the probability of type II error under the alternative pl 08 is about 23 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Small Sample tests 0 These are test procedures for proportions when the sample size n is small They are based directly on the binomial distribution rather than the normal approximation 0 Consider the alternative hypothesis Ha p gt p0 and let X be yet again the number of successes in the sample size n For a test level 05 we find the rejection region from PX 2 cwhen X N Binnp0 1 PX g c 1 when X N Binnp0 1 Bc 1np0 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 a It is usually not possible to find an exact value of C in this case the usual way out is to use the largest rejection region of the form 6 C 1 n satisfying the bound on the Type I error a To compute the Type II error for an alternative pl gt 190 we first note that X N Binnpl if the alternative is true Then mpl PX lt cwhen X N Binnpl Bc 1npl Note that this is a result of a straightforward binomial probability calculation Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Example 0 A builder claims that heat pumps are installed in 70 of all homes being constructed today in the city of Richmond VA Would you agree with this claim if a random survey of new homes in this city shows that 8 out of 15 had heat pumps installed Use a 01 level of significance 0 H0 19 07vs Ha p lt 07with Oz 010 o The test statistic is X N Bin07 15 Aug 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 a We have a 8 and npo 1507 105 Thus we must find csuch that PX 2 c1 Bc 11507 01 for X N Bin07 15 It is easy to check that the rejection region will be 137 147 15 Aug 2006

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