Statistical Inference PDF

STATISTICAL INFERENCE I Generally: Drawing conclusion about populations from samples, by means of statistical evidences Specifically: we want to estimate and draw conclusion about value of parameters of o...

STATISTICAL INFERENCE I Generally: Drawing conclusion about populations from samples, by means of statistical evidences Specifically: we want to estimate and draw conclusion about value of parameters of one or more populations using HYPOTHESIS TESTING populationparameters mean M population parameter population p  From given situation, data samples & parameters I Estimate/Create assumption statement (Hypothesis) Prove whether the statement is reasonable/acceptable or not using statistical analysis. A test to prove whether HYPOTHESIS the statements are TESTING reasonable or not. TYPES OF HYPOTHESIS TESTING (DEPENDING ON TYPES/NUMBER OF SAMPLES AND DATA) A ONE SAMPLE HYPOTHESIS TESTING deals with ONESAMPLE DATASET TWO SAMPLE HYPOTHESIS TESTING B deals with 1 TWO INDEPENDENT DATASET Eg 2 different companies TWO DEPENDENT DATASET PAIRED Ii Eg Before After Pre post c ANOVA deals with more than TWO INDEPENDENT DATASET D CHI SQUARE deals with NOMINAL CATEGORICAL DATASET  HYPOTHESIS TESTING A ONE SAMPLE HYPOTHESIS TEST Step 1 Hypothesis Development Hi create two hypothesis statement Ho Null Hypothesis describeexists populationparameters Alternate Hypothesis mean M or proportion µ opposite of Ho Three ways to write thestatements depending on situations and questions Tiffani tianya.i.it i ig E instrument Ho M 8 H M 8 Example 2 at least 80 oforders will be distributed Ho P 0.8 H P 0.8 waiting time is 15minutes y Etmihftghe.ir mean waiting time is different from Ho M 15 H M 15 step ftp.t.it b tw i aitettest Example1 checkthesign of H H M 78 indicatesupper tailed 11 Example checkthesign of H H p 20 8 indicates lower tailed Example checkthesign of H H M 15 indicates two tailed Titision  α 0.05 Step 3 SelectSignificanceLevel Definition a measure or probability of decision to reject Ho denoted as α usually 4 0.05 is selected NOTE: HYPOTHESIS TEST LEADS TO FOUR POSSIBLE DECISIONS: 1) H0 is true. Hypothesis test fails to reject H0 2) H0 is false. Hypothesis test rejects H0 3) H0 is true, but the test rejects H0 (Type I error, probability = α) 4) H0 is false, but the test fails to reject H0 (Type II error, probability = ) β  Step 4 calculate critical value and p value A critical value Tomputeavalue to beused to decidewhether to reject Ho vane 1 Compute test statistic and critical 2 2 5hpm If is known use If Δ is unknown use t Iff compute critical t distribution value usingexcel function tail Upper T INV 1 α n 1 Hypothesis formean α n1 Lowertail T INV Ho M Twotail T INV 2T A N 1 critical z value using excelfunction Hypothesis forproportion compute NORM S INV 1 α Ho p EE ft 2 Plot the rejection region criticalvalue  B P value t and s obtainedfrom Excel Function Compute p value using tail Upper I T DIST t n 1 TRUE Hypothesis formean Lowertail T DIST t n 1 TRUE Ho M Twotail T.DIST.IT tint if t o T Dist 2T tn1 if t 0 Hypothesis forproportion Lowertail Norm S DIST Z TRUE Ho P two tail 2 1 NORM S DIST Z TRUE if 270 2 1 NORM 5 DIST 2 TRUE if 20  critical value Step 5 Decision p value 1 Critical value If 2 or t obtained from fallsunder reject rejection region then falls under non rejection region then failovejectH 2 P value compare theobtained p value with 0.05 If p value 0.05 reject Ho If pvalue 0.05 failed to reject Ho ONE-SAMPLE HYPOTHESIS TESTING Example 5.1 A fast food restaurant claims that customer's order will be served within 8 minutes. The manager received complaints and collected a sample of 50 customers waiting time. Use hypothesis testing , to provide evidence that the sample mean waiting time supports the claim made by the restaurant. Ho M 8 H M 8 Based on H M 8 upper tail significance level 0.05 t stat value as 6 is unknown data 5,11 computed using given in Excel 0.105 i critical value approach compute critical value using Excel function forupper tail T INV 1 α n 1 T INV 1 0.05 50 1 TINV 0.95 49 1.677 ii pvalueapproach function for upper tail compute pvalue using Excel I T DIST t n 1 TRUE 1 T DIST 0.105 49 TRUE 20.4584  if critical value 0.105 fallsunder non rejection region failed to reject Ho ii p value 0.4584 4 0.05 failed to reject Ho i conclusion There is sufficient evidence to support claim made restaurant that the sample by mean waiting time is within 8 minutes Example 5.2 Ho H M N 15 15 IIII 4 0.05 compute t 4.0920 2,1 i critical value Excel function of twotailed since H M is indicates my T.NU 2T 0.05 39 2.023C ii pvalueapproach function for two tail compute pvalue using Excel T DIST 2T 4.092 39 00020 0 111s i t 4.092 falls in rejection region reject Ho ii 0.0002 4 0.05 reject Ho There is sufficient evidence that themean waiting time is not 15 minutes Example 5.3 proportion of iiitied If Ho p Hi p as 2 0.05 t stat using 2 7 12 1.82574 i critical z value using Excel function NORM S INV 1 α 1.64485 Based on H pcos horse iiÉr tail and 1.64485 ii pvalueapproach functionforonetail z test compute pvalue using Excel Norm 5 DIST C 1.826 TRUE 0.0339 Me 1.64485 2 1.82574 falls under rejection region reject Ho ii 0.0339 4 0.05 reject Ho The claim made by company that at least 80 of all Orders reached customers within 3 days is not true  HYPOTHESIS TESTING B TWO SAMPLE HYPOTHESIS TEST TWO INDEPENDENT SAMPLES Step 1 Hypothesis Development i create two hypothesis statement Ho Null Hypothesis Alternate It Hypothesis ii Three ways to write thestatements depending on situations and questions velage score for class A is more than B Ho Ma MB O H MA MB 0 class A is less than B go score for Ho Ma MB 0 H MA MB C O g's score for class A is no different with B era Ho MA MB 0 It MA MB 0  Determine the one tailed or twotailedtest step 2 Determine the upper tailed or lower tailed test Eemple1 check thesign of H Hi Ma Ma 70 indicatesupper tailed freight Example 2 checkthesign of H H MA up 0 indicates lower tailed Example check thesign of H H Ma M 0 indicates two tailed fittson Important 1 t stat if t is positive value uppertail is negative value lowertail 2 Ipt p From if t is positiveorzero pvalue stated is the pvalueforuppertail to obtain pvalueforlowertail 1 pvalue From if t is negative pvalue stated is the pvalueforlowertail to obtain pvalueforuppertail 1 pvalue 3 criticalvalue criticalvalue intheoutputis foruppertail simply putnegativesign forlowertail  level 9 0.05 Step 3 Significance Step 4 calculate critical value α p value Use Excel to perform Data Analysis Choose Appropriate Analysis Tools Two Equal Variances t test Samples Assuming ttest Two Samples Assuming Unequal Variances Most practical 2 test Twosamples for Means criticalvalue Steps Decisions pvalue Decision toreject Ho is similar to ONESAMPLE if fallsinrejection region Evalue with criticalvalue I sttorejectHo if fallsin nonrejection region pvalue with 4 0.05 rejectHo if p 0.05 failstorejectHo if p 0.05 TWO-SAMPLE HYPOTHESIS TESTING (Independent) ftp.fttonthetc Identitiesof LED Technologies are longer Example 5.4 2191 Ho Mi Ms 0 α 0.05 H Mi Ma 70 Data Analysis in Excel i select t test Twosample assuming unequal variance ii 4 0.05 Interpret result Note Based on H M Ma70 t stat 5.13 uppertail Pvalue 0 00012 of positive uppertail t critical 1.78 positive positive Decision t stat falls inrejection region reject Ho aol.ES reject Ho pvalue Leadtime for LED Technologies is statistically longerthan that of sks Enterprise  HYPOTHESIS TESTING B TWO SAMPLE HYPOTHESIS TEST PAIRED SAMPLES dependent fromsameelements step 1 Hypothesis Development i create hypothesis statement for Twopairedsamples d mean ofthe Ho Md 0 paired differences H Ma O H nia o or H Md 0 dataset of before and aftersales Seller wants to improve and increase the sales improve means after before Ho Md 0 Hi Md 70 step B ftp.tt eiitiiti aitettest Example1 checkthesign of H H Md 70 indicatesupper tailed S 11 Example checkthesign of H H Md C O indicates lower tailed Example checkthesign of H H Md 0 indicates two tailed Titision level α 0.05 Step 3 Significance Step 4 Calculate critical value α p value Use Excel to perform Data Analysis 1 Choose t test Pairet Two sample forMeans 2 Select the data Difference 3 Fill in O in Hypothesized Mean 4 Check Labels 5 Fill in Alpha 0.05 In excel Data Analysis choose the variable that is more than as variable 1 Interpreting the output similar as TWO SAMPLES INDEPENDENT criticalvalue Step5 Decision pvalue similar as TWO SAMPLES INDEPENDENT TWO-SAMPLE HYPOTHESIS TESTING (Paired) c Example 5.5 81in anti c p ftp.ifeng eeniteiaiseiniabour or Ho Md 0 H Md 0 11 in 1 4 0.05 Data Analysis in Excel i select t test Paired Two samples forMoan ii Hypothesized mean difference 0 iii Label iv Alpha 0.05 Interpret result Note t stat 8.512 negative lowertail Pvalue 0 0000018 t critical to negative since it 1.795 change islowertail Decision t stat falls inrejection regionarejoct.it 1.795 2 9 0.05 reject Ho pvalue New picking proless does improve the picking rateof pickers  F- TEST To test for equality of variances between two samples. Step 1 Hypothesis Development step2 upffatedTwo tailed Ho 62 63 Step3 α 0 05 Hi 6 65 Step44 2 Excel Data analysis tool choose F test the sample forvariances i Select variable variablewith largersample variance isassigned as Variable l ii uncheck labels I iii 9 0.05 step 5 3 Interpret Results Decisions i F value positive uppertail ii p value iii F critical i Use F value F critical to plot rejection region i compare pvalue with D 0.05 p value h rejectHo a D fail to pvalue reject Ho Emp step 1 Ho 62 6 Equalvariance It G 6 unequal Variance Steps upper level Two tailed steps 9 0.05 Excel Data Analysis F test stop 4 2 output F 4 0.0286 p F critical 3.6 5 Step 3 Interpreting Results FFF 1111 F intical 4 3.6 RejectHo i F falls in rejection reject ii p value 0.0286 C 4 0.05 i reject Ho ANOVA i Populations follow normal distributions Assumptions variance ii Populations have equal iii Populations are independent Hypothesis Step 1 Ho M M Ms H Not all three means are equal Determine the one tailed or twotailedtest step 2 Determine the upper tailed or lower tailed test level α 0.05 step 3 Significance critical value Pvalue step 4 449glata.I atisatiy.it Decisions P critical value using F statistics step 5 level p value using significance CHI-SQUARE distributed Population are not normally i Assumptions it For categorical data only Hypothesis Step 1 Ho Two categorical variables are independent dependent Hi Two categorical variables are step ftp.tt feaitiiti aitettest level 9 0.05 Step 3 Significance critical value Pvalue step 4 449glataatg sqnqij.is Decisions P critical value using chi square step 5 level p value using significance