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# Introduction to Statistics

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Introduction to Statistics Hypothesis Testing Valerie Chu Ph D Summer 2004 Hypothesis Testing Hypothesis testing is a process of making decision or inference from a set of sample data for evaluating claims about a population For example you try to investigate whether an individual s blood pressure is higher under stress From your experiments someone gets higher blood pressure after running or riding an exercise bike But others may not How do you use your data to support your claim You must go through statistical hypothesis testing to make a conclusion about the truth of your claim Null Hypothesis The Null Hypothesis is denoted by H0 and is assumed to be true until sufficient evidence is obtained to warrant its rejection For example there will be no increase on blood pressure after an exercise Alternative Hypothesis The Alternative Hypothesis is denoted by H1 and is opposite of Null Hypothesis The researcher wants to prove it For example there will be increase on blood pressure after an exercise Making Decision p value then reject H0 p value then fail to reject H0 where is the specified significance level e g If p value 0 0431 0 05 then reject H0 at 5 significance level That is under null hypothesis there is 4 31 chance of getting the test statistic value Since we allow 5 error margin i e significance level we reject the null hypothesis based upon the test statistic value 0 431 0 0 05 1 Project Comparison of normal BP with BP under cold stress and exercise induced stress Objective People with potential to develop hypertension and high blood pressure were shown to have their blood pressure elevate under stress This simulated stress induced by cold temperature and exercise would help to identify subjects who could develop hypertension later in life Methods Participants would be subjected to cold stress by immersing their hand in ice cold water for a few minutes and then their BP would be recorded and compared with BP recorded under normal condition Subjects would also be asked to ride an exercise bike for 10 15 minutes and their BP would be recorded and compared with BP recorded under normal condition Example 1 Type the collected data into Excel Sheet as below BP Normal Bp Exercise 130 126 160 167 186 197 153 135 123 132 132 150 143 142 147 169 155 134 122 124 H0 BP Exercise BP Normal H1 BP Exercise BP Normal 2 Highlight the columns of BP Normal and BP Exercise 3 Click Data Analysis from Tool menu 4 In Data Analysis select t Test Paired Two Sample for Means and click the OK button 2 5 In the t Test Paired Two Sample for Means dialog box select the Variable 1 Range Variable 2 Range Hypothesized Mean Difference Labels Output and Range and click the OK button Explanation of your analysis results t Test Paired Two Sample for Means Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P T t one tail t Critical one tail P T t two tail t Critical two tail BP Normal Bp Exercise 145 1 147 6 387 2111111 546 9333333 10 10 0 799759053 0 9 0 561754415 0 293998702 1 833113856 0 587997404 2 262158887 Since p value 0 293998702 0 05 fail to reject null hypothesis That is the exercise does not increase blood pressure according to the data from the above example 3 ANOVA When an F test is used to test a hypothesis concerning the means of three or more populations the technique is called analysis of variance commonly abbreviated as ANOVA That is the hypotheses will be set up as below H 0 1 2 n H 1 At least one mean is different from the others Example from Bluman s book A researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure The subjects are randomly assigned to three groups the first group takes medication the second group exercises and the third group follows a special diet After four weeks the reduction in each person s blood pressure is recorded At 0 05 we test the claim that there is no difference among the means The following table shows the data Medication Exercise Diet 10 12 9 15 13 6 8 3 0 2 5 9 12 8 4 1 Enter data and set up the hypotheses H 0 1 2 3 null hypothesis H 1 At least one mean is different from the others 2 Select Tools Data Analysis and choose Anova Single Factor 4 3 Enter the data range for the three columns Set an alpha level Select a location for output and click OK 4 The results are shown Anova Single Factor SUMMARY Groups Medication Exercise Diet Count 5 5 5 Sum 59 19 38 ANOVA Source of Variation Between Groups Within Groups SS 160 1333 104 8 df 2 12 Total 264 9333 14 Average Variance 11 8 5 7 3 8 10 2 7 6 10 3 MS F P value 80 06667 9 167939 0 003831 8 733333 F crit 3 88529 5 Make the decision The decision is to reject the null hypothesis since the p value is 0 003831 which is less than 0 05 the alpha value 6 Summarize the results There is enough evidence to reject the claim and conclude that at least one mean is different from the others 5 Two Way Analysis of Variance If ANOVA involves two independent variables it is called two way analysis of variance Project Rate of diffusion of methylene blue and commassie blue dye under three physical conditions Objective To learn about the affect of temperature on the diffusion of dye molecules in a liquid medium Heat energy excites molecules and increases the kinetics of molecular motion In this exercise participants would use icecold water boiling water and normal tap water and record the time it takes for the complete diffusion of the dye molecules in water under the above physical conditions Example 1 Enter the following data set Methylene Blue Commassie Blue Tap Water 10 15 17 12 18 17 Ice Water 8 5 9 10 6 13 Boiling Water 15 18 12 18 19 15 2 Select Tools Data Analysis and choose Anova Two Factor with Replication 6 3 Enter the data range including labels Also enter the number of rows per sample Select output range and click OK 4 The results are shown below ANOVA Source of Variation Sample Columns Interaction Within Total SS 201 3333333 20 05555556 0 444444444 106 6666667 df 2 1 2 12 328 5 17 MS F P value F crit 100 6667 11 325 0 001725 3 88529 20 05556 2 25625 0 158926 4 747221 0 222222 0 025 0 975361 3 88529 8 888889 5 Make the decision The decisions are a to reject the null hypothesis H0 There are no different effects among ice cold water boiling water and normal tap water since p value 0 001725 0 05 b to fail to reject the null hypothesis H0 There are no different effects between methylene blue …

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