Data Analysis and the Creation of a Statistical Test

Test Scores: 95, 92, 90, 90, 83, 83, 83, 74, 60 and 50.


The mean for the test scores listed above is 80. This figure wasarrived at by first establishing the number of data points in thesample (often denoted as ”N”), before proceeding to find the sum total in the setprovided. Dividing the sum by “N” (95+92+90+90+83+83+83 +74+60+ 50/ 10) therefore equaled 80 as the mean for the test scores.
The median for the test scores provided above is 83. This is because “median”, in any grouped data, refers to the figure separating the first half of the samplefrom the second (Rubin, 2012). It is computed by arranging the data set in ascendingorder before choosing the number(s) that fall in the middle of theset.  It is, however, worth noting that this only applies to situations where anodd number falls in the middle of the data set. Whenever even numbersappear at the center, their average then becomes the median. In thisparticular case, it happened to be 83 +83 which would then be dividedby 2 to give 83 as the median.
The mode in any data set refers to a number that appears morefrequently. This happens to be 83 since it appears thrice.
The range in this data set is 45. It is computed by finding thedifference between the highest and lowest numbers listed. In thiscase, the sample size is 10, with the highest figure being 95 and thelowest 50. Subtracting the lowest figure from the highest (95-50) therefore resulted in 45 as the range.
The variance in this data set is 214.7. It is calculated by findingthe averaged square differences of the mean. Beginning with the simpleaverage in the data set for each individual number, it would then besubtracted from the mean before finally being squared (50²+60²+74²+83²+83²+83²+90²+90²+92²+95² =65,932 – 8002). The last stepentails establishing the average from the squared differences. Dividing 1,932 by 9 would give the variance; 214.7
The Standard Deviation for the data set provided above is 14.65. It iscalculated by finding the square root of the established variance.This is done by first finding the sum total in the data set, which is800. It is then squared resulting in 640,000.The next step would beto divide this figure by 10, which is the total number of figures inthe sample, resulting in 64,000. The next step is finding the squaresof all individual numbers in the data set, before adding them all up:50²+60²+74²+83²+83²+83²+90²+90²+92²+95² = 65,932. The next step wouldbe to subtract 64,000 from this figure, which would result in 1,932 asa final figure. Then, one would have to subtract 1 from the totalnumber of samples in the data set (10) to ultimately end up with9.Dividing 1,932 by 9 would give the variance, which is 214.7. Thestandard deviation hence hails from the square root of this figure; 14.65.

It would be fitting to conduct a One-Sample T-Test since it would givethe differences between the mean score and the various continuitylevels after normal distribution of data. Moreover, this test willalso provide an accurate comparison of the mean from one sample.
When inter-relating results from a population with a mean of 70 and asignificance of 0.05,the goal was to establish whether the driverswould drive at the same speed as they would at 70 mph since it is theknown mean for the population. The sample population drivers are 60, 60,55, 50, 45, 45, 55, 80, 80, 70, 70, 70, 75, 75, 65 and 60. From the test, theT-Value happened to be -2.28 while 0.037 represented the P-Value whichI then rounded up to 0.04. The significant level of the data was setat 0.05 which was greater than “p”. Thus, it would be appropriate toreject the null hypothesis as there is evidence of considerablediscrepancies between the sample mean and that of 70. Such data isvital during occasions when a researcher would want to find out whether or not a sample population differs from the general population
in a significant way.

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