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Analyzing the Effectiveness of Vaccines and Understanding Statistical Significance in Research

Discusses vaccines and statistical significance.

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    Analyzing the Effectiveness of Vaccines and Understanding Statistical Significance in Research 1. A group of researchers conducted an experiment to determine which vaccine is more effective for preventing getting the flu. They tested two different types of vaccines: a shot and a nasal spray. To test the effectiveness, 1000 participants were randomly selected with 500 people getting the shot and 500 the nasal spray. Of the 500 people were treated with the shot, 80 developed the flu and 420 did not. Of the people who were treated with the nasal spray, 120 people developed the flu and 380 did not. The level of significance was set at .05. The proportion of people who were treated with the shot who developed 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 and nasal spray used for preventing nasal spray? To examine the difference of effect we conduct test based on given observation i.e., n 1 = 500 , n 2 = 500 ,p 1 = 0.16 , p 2 = 0.24 , alpha = 0.05 Our Hypothesis is defined as Null Hypothesis : s hot spray i s equal effective than nasal spray Vs Alternative Hypothesis : shot spr ay is more effective than nasal spray Since sample sie are large enough ( >30) To conduct this testing we use Z test with p as pooled estimate of proportion = 0.20 Z= (p 1 – p 2 ) / sqrt(p*1 - p *(1/n 1 + 1/n 2 ) = - 0.08 /0.025 = - 3.2 P( Z < - 3.2) = 0.0008 P 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. We can say that there is significance difference between shot and nasal spray. We can say that shot is more effective than the nasal spray. Since n * p i > 5 ,thus we can say that our test is ok We assume our alternative hypothesis is shot is more effective than nasal spray based on sample proportion but it can be viceversa also. So our result does not provide enough evidence to support alternative hypothesis because in other alternative calculation remain same and it gives same z with p value Since proportion follow binomial distribution to approximate it to n ormal we need large sample size , on a thumb rule say n > 100 . We have n =500 therefore we can say that this sample is appropriate for study the effect. After the calculation we can not say confidently that which one is better a short or nasal spray? , we are assuming on thumb rule that data follow normal distribution but we have not evidence that it actually follows approximated normal distribution. These are the possible limitations of the study We would conduct a follow up study by paired test because these are independent samples ,means every person has different stamina ,so better option is to conduct a pair test in which same person given both type of vaccines on a certain time. T his study used chi - square test for contingency table 2 X 2. S tatistical significance is mathematical - it comes from the data (sample size) and from your confidence (how confident you want to be in your results). Practical significance is more subjective and is based on other factors like cost, requirements, program goals, etc. For example, suppose a college transition program is developing a survey to assess student preferences for certain types of post secondary options (i.e. community colleges, public senior colleges, private institutions, military service, full - time work, etc.). Often the first question programs ask is how many responses are needed to get “statistically significant” results. That is where the confusion starts; that question only makes sense in the context of a statistical hypothesis test. A survey may involve many hypotheses that we want to test. A statistical hypothesis test requires both a hypothesis: women enroll in community colleges more than men, and a test statistic: the percent of women who enroll in community colleges minus the percent of men who enroll in community colleges. Now we can ask if the test statistic (the difference between the two percentages) is “statistically significant.” That’s a legitimate question. A more meaningful question might be whether the difference is “practically significant.” 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 - based services toward women than men?) Statistical significance depends on the sample size. A difference of 3% (58% for women minus 55% for men) can be statistically significant if the sample size is big enough, but it may not be practically significant. 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 a decision to focus more resources in one direction) but if the difference isn’t statistically significant (that depends on sample size) then you can’t be sure the difference you see (30%) is real, so you either need to 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 found the 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 r 2 = 0.56 which simply means iq level accounted 56 % in gpa and rest are the others factor, i.e., time spent by student on study , education level , family background etc.This correlation shows that as level of iq increases , gpa will also increase means high iq level causes high gpa Not only iq lev el is responsible for gpa level, there are others factors also like performance, education level , family background, economic condition, time given by student to study etc variables might have i nfluence on the relationship. C orrelation only shows relation - strong or week, it can' t use to quantify the outcome. F or example if iq level will be increase by x(say), then ho w much gpa level will increase?. S o for this reg ression (causation)is needed. Mainly there are six variables affecting size of correlation (a) The amount of variability in either variable, X or Y (b) D ifferences in the shapes of the two distributions, (c) Lack of linear ity in the relationship between X and Y (d) T he presence of one or more “outliers” in the data set (e) C haracteristics of the sample used for the calculation of the correlation and (f) M easurement 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 the exact linear relation between variables. 3. A researcher has recorded the reaction times of 20 individuals on a memory assessment. The following table indicates the individual times: 2.2 4.7 7.3 4.1 9.5 15.2 4.3 9.5 2.7 3.1 9.2 2.9 8.2 7.6 3.5 2.5 9.3 4.8 8.5 8.1 Solution By using excel we get Mean 6.36 Standard Error 0.758995181 Median 6.05 Mode 9.5 Standard Deviation 3.394329637 Sample Variance 11.52147368 Kurtosis 0.578947484 Skewness 0.77523285 Range 13 Minimum 2.2 Maximum 15.2 Sum 127.2 Count 20 Largest(1) 15.2 Smallest(1) 2.2 Confidence Level(95.0%) 1.58859517

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    Mean of data is 6.36 which means on an average persons have 6.36 seconds reaction time, but there is outlier in the data ,therefore mean is not a good measure of central tendency ,we use median as measure of central tendency to measure the average reaction time. Suppose we omit the outlier’s i.e., 15.2 then our descriptive statistics are Mean 5.894736842 Standard Error 0.632107479 Median 4.8 Mode 9.5 Standard Deviation 2.755292621 Sample Variance 7.591637427 Kurtosis - 1.795892701 Skewness 0.060172236 Range 7.3 Minimum 2.2 Maximum 9.5 Sum 112 Count 19 Largest(1) 9.5 Smallest(1) 2.2 Confidence Level(95.0%) 1.328008534 We can observe that there were no outliers in the data hence we can say that average response time is mean response time which is 5.89 and it is less from previous mean . Group 1 data are Group 2 data are 2.2 2.5 2.7 2.9 3.1

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    Descriptive statistics for different groups are Group 2 Mean 9.24 Standard Error 0.707452 Median 8.85 Mode 9.5 Standard Deviation 2.237161 Sample Variance 5.004889 Kurtosis 6.86883 Skewness 2.443916 Range 7.9 Minimum 7.3 Maximum 15.2 Sum 92.4 Count 10 both 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 group After doubling the data new gropus are formed as 3.5 4.1 4.3 4.7 4.8 7.3 7.6 8.1 8.2 8.5 9.2 9.3 9.5 9.5 15.2 Group1 Mean 3.48 Standard Error 0.297695 Median 3.3 Mode #N/A Standard Deviation 0.941394 Sample Variance 0.886222 Kurtosis - 1.5709 Skewness 0.190223 Range 2.6 Minimum 2.2 Maximum 4.8 Sum 34.8 Count 10

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    Group 1 Group 2 2.2 7.3 2.5 7.6 2.7 8.1 2.9 8.2 3.1 8.5 3.5 9.2 4.1 9.3 4.3 9.5 4.7 9.5 4.8 15.2 2.2 7.3 2.5 7.6 2.7 8.1 2.9 8.2 3.1 8.5 3.5 9.2 4.1 9.3 4.3 9.5 4.7 9.5 4.8 15.2 Now descriptive statistics for both different goups are Group 1 Group 2 Mean 3.48 Mean 9.24 Standard Error 0.204888 Standard Error 0.486902 Median 3.3 Median 8.85 Mode 2.2 Mode 9.5 Standard Deviation 0.916285 Standard Deviation 2.177493 Sample Variance 0.839579 Sample Variance 4.741474

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