Analyzing the Effectiveness of Vaccines and Understanding Statistical Significance in Research

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Analyzing the Effectiveness of Vaccines and Understanding StatisticalSignificance in Research1.A group of researchers conducted an experiment to determine which vaccine is more effective forpreventing getting the flu. They tested two differenttypes of vaccines: a shot and a nasal spray. To testthe effectiveness, 1000 participants were randomly selected with 500 people getting the shot and 500the nasal spray. Of the 500 people were treated with the shot, 80 developed the flu and 420 did not. Ofthe people who were treated with the nasal spray, 120 people developed the flu and 380 did not. Thelevel of significance was set at .05. The proportion of people who were treated with the shot whodeveloped the flu = .16, and the proportion of the people who were treated with the nasal spray was .24.The calculated p value = .0008.We have to research Is there any significant difference between two typres of vaccine i.e, a shot andnasal spray used for preventing nasal spray?To examine the difference of effect we conduct test based on given observation i.e.,n1= 500 , n2= 500 ,p1= 0.16 , p2= 0.24 , alpha = 0.05Our Hypothesis is defined asNullHypothesis : shot spray is equaleffective thannasal sprayVsAlternativeHypothesis : shot spray is moreeffective thannasal spraySince sample sie are large enough ( >30)To conduct this testing we useZ test with pas pooled estimate of proportion = 0.20Z= (p1–p2) / sqrt(p*1-p *(1/n1+ 1/n2)=-0.08 /0.025=-3.2P(Z<-3.2) =0.0008P value = 0.0008 < 0.05(significance level)

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Thus on basis of p value we can say that we can reject null hypothesis at 5 % of level of significance. Wecan say that there is significance difference between shot and nasal spray.We can say that shot is moreeffective than the nasal spray.Since n * pi> 5 ,thus we can say that our test is okWe assume our alternative hypothesis is shot is more effective than nasal spray based on sampleproportion but it can be viceversa also. So our result does not provide enough evidence to supportalternative hypothesis because in other alternative calculation remain sameand it gives same z with pvalueSince proportion follow binomial distribution to approximate it to normal we need large sample size, ona thumb rule say n > 100 . We have n =500 therefore we can say that this sample is appropriate for studythe effect.After the calculation we can not say confidently that which one is better a short or nasal spray? , we areassuming on thumb rule that data follow normal distribution but we have not evidence that it actuallyfollows approximated normal distribution. These are the possible limitations of the studyWe would conduct a follow up study by paired test because these are independent samples ,meansevery person has different stamina ,so better option is to conduct a pair test in which same person givenboth type of vaccineson a certain time. This study used chi-square test for contingency table 2 X 2.Statistical significance is mathematical-it comes from the data (sample size) and from your confidence(how confident you want to be inyour results). Practical significance is more subjective and is based onother factors like cost, requirements, program goals, etc.For example, suppose a collegetransition program is developing a survey to assess student preferencesfor certain types of postsecondary options (i.e. community colleges, public seniorcolleges, privateinstitutions, military service, full-time work, etc.). Often the first question programs ask is how manyresponses are needed to get “statistically significant” results. That is where the confusion starts; thatquestion only makes sense in the context of a statistical hypothesis test. A survey may involve manyhypotheses that we want to test. A statistical hypothesis test requires both a hypothesis: women enrollin community colleges more than men, and a test statistic: the percent of women who enroll incommunity colleges minus the percent of men who enroll in community colleges. Now we can ask if thetest statistic (the difference between the two percentages) is “statistically significant.” That’s a legitimatequestion. A more meaningful question might be whether the difference is “practicallysignificant.”Practical significance: A calculated difference is practically significant if the actual difference it is

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estimating will affect a decision to be made. (Should the program focus more community college-basedservices toward women than men?)Statistical significance depends on the sample size. A difference of 3% (58% for women minus 55% formen) can be statistically significant if the sample size is big enough, but it may not be practicallysignificant. 3% hardly seems big enough to warrant focusing on one group of students over the other.A difference of 30% (65% for women minus 35% for men) may be practically significant (i.e., warrant adecision to focus more resources inone direction) but if the difference isn’t statistically significant (thatdepends on sample size) then you can’t be sure the difference you see (30%) is real, so you either needto get more data or treat the two groups as the same.2.A researcher has investigated the relationship between IQ and grade point average (GPA) and foundthe correlation to be .75.r = 0.75, 0.7 < r <1 is strong positive correlation between iq level and gpa.Basically r=0.75 gives r2= 0.56 which simply means iq level accounted 56 % in gpa and rest are the othersfactor,i.e.,time spent by student on study , education level , family background etc.This correlationshows that as level of iq increases ,gpa will also increase means high iq level causes high gpaNot only iq level is responsible for gpa level,there are others factors also like performance, educationlevel, family background, economic condition, time given by student to study etcvariables might haveinfluence on the relationship.Correlation only shows relation-strong or week, it can't use to quantify the outcome. For example if iqlevel will be increase by x(say), then how much gpa level will increase?. So for this regression(causation)is needed.Mainly there are six variables affecting size of correlation(a) The amount of variability in either variable,XorY(b) Differences in the shapes of the two distributions,(c) Lack of linearity in the relationship between Xand Y(d) The presence of one or more “outliers”in the dataset(e) Characteristics of the sample usedfor the calculation of the correlationand(f) Measurement error.Correlation can not be used for predicting gpa, because correlation only tells about linear relationship

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not what linear relationship occurs.for predicting Gpa regression will be used because it quantify theexact linear relation between variables.3.A researcher has recorded the reaction times of 20 individuals on a memory assessment. The followingtable indicates the individual times:2.24.77.34.19.515.24.39.52.73.19.22.98.27.63.52.59.34.88.58.1SolutionBy using excel we getMean6.36Standard Error0.758995181Median6.05Mode9.5Standard Deviation3.394329637Sample Variance11.52147368Kurtosis0.578947484Skewness0.77523285Range13Minimum2.2Maximum15.2Sum127.2Count20Largest(1)15.2Smallest(1)2.2Confidence Level(95.0%)1.58859517

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Mean of data is 6.36 which means on an average persons have 6.36seconds reaction time, but there isoutlier in the data ,therefore mean is not a good measure of central tendency ,we use median asmeasure of central tendency to measure the average reaction time.Suppose we omit the outlier’s i.e.,15.2 then our descriptive statisticsareMean5.894736842Standard Error0.632107479Median4.8Mode9.5Standard Deviation2.755292621Sample Variance7.591637427Kurtosis-1.795892701Skewness0.060172236Range7.3Minimum2.2Maximum9.5Sum112Count19Largest(1)9.5Smallest(1)2.2Confidence Level(95.0%)1.328008534We can observe that there were no outliers in the data hence we can say that average response time ismean response time which is 5.89 and it is less fromprevious mean .Group 1 data areGroup 2 data are2.22.52.72.93.1

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Descriptive statistics for different groups areGroup 2Mean9.24Standard Error0.707452Median8.85Mode9.5StandardDeviation2.237161Sample Variance5.004889Kurtosis6.86883Skewness2.443916Range7.9Minimum7.3Maximum15.2Sum92.4Count10both groups are differ on every variables i.e., mean,median standard deviation,kurtosis etc.there is outlier in second group it increases the average response time for the groupAfter doubling the data new gropus are formed as3.54.14.34.74.87.37.68.18.28.59.29.39.59.515.2Group1Mean3.48Standard Error0.297695Median3.3Mode#N/AStandardDeviation0.941394Sample Variance0.886222Kurtosis-1.5709Skewness0.190223Range2.6Minimum2.2Maximum4.8Sum34.8Count10

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Group 1Group 22.27.32.57.62.78.12.98.23.18.53.59.24.19.34.39.54.79.54.815.22.27.32.57.62.78.12.98.23.18.53.59.24.19.34.39.54.79.54.815.2Now descriptive statistics for both different goups areGroup 1Group 2Mean3.48Mean9.24Standard Error0.204888Standard Error0.486902Median3.3Median8.85Mode2.2Mode9.5StandardDeviation0.916285StandardDeviation2.177493Sample Variance0.839579Sample Variance4.741474

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