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¶ … median and the measure of dispersion is the standard deviation. The mean and standard deviation were computed using the relevant formula.
Formula for the mean =?X/N
Group A
Group B
Based on the analysis of the statistical measures employed, I would choose group B. The rational for my choice is as follows: While the means of both groups are the same; thus suggesting that the groups would be equally good. The median for the group B. is closer to its mean. The median is a measure of the physical center of the distribution; extremely high or low scores do not influence it. The median for group B. is only 0.2 points higher than the mean. This indicates that there are fewer scores within group B. with the potential to influence the mean excessively. When a distribution contains scores that are much higher or lower than the mean they can inflate or deflate the value of the mean. In the group A, the median is lower than the mean by 3.2 points. It is possible to infer from that that there are low scores in the distribution that are deflating the value of the mean. These low scores represent persons whose handling skills for fragile material are weak and are a potential risk to the company. In group B, the handling skills of the group members are very similar.
Group B. also had the lower standard deviation 12.43 as compared to 22.71. The standard deviation is a measure of the difference in skill between the group, the larger the standard deviation, then the greater the quantum of difference between the scores. Since you must employ the entire group, it is an imperative that the chosen group has the smallest differences in their skill level. If the differences in skill level are too great, what may occur is that you have persons who are highly skilled doing a great job, but persons of poor skill doing a terrible job undermine this.
These measures provide an adequate tool for a statistical assessment of the two groups of individuals. AT-test would be a useful addition to the available tools used to assess the difference between the groups. The T-test would determine if the observed difference was significant or an artifact of chance.
Looking at the variability of each group is important because it gives the decision-maker an indication of how wide is the spread of scores of each worker in the group. That is, through measures of variability, the decision-maker is informed of just how varied the manual dexterity scores of the workers are, since it would be most efficient for him/her to have workers who at least have the same levels/scores
median and the mode. The mean is the average of the total set, the median is the middle of the total set and the mode is the most frequently reported number of the total set. The median can be found by literally writing the numbers in a string, and then finding the number that sits exactly in the middle. When measuring family wealth in order to make political decisions,
GPA is interval level data one can report all measures of central tendency and dispersion (Creswell, 2012). In the original data case 38 had a GPA reported as 9.00. This value is of course impossible and since there were no missing values defined in the data set this case was treated as missing data. If used as reported it would have resulted in a much larger standard deviation and
A standard deviation is so "standardized" that, generally speaking, 70% of the data points will fall within one standard deviation, and 90% of the data points will fall within two standard deviations. Standard deviation is one of the most common ways for researchers to describe data as it forms a basis for comparing data from study to study. Measures of central tendency are most effective at describing data that shows
46). The third measure of central tendency is the mode. Despite it being the last option of consideration by many analysts, it is a mostly utilized measure. The mode represents the most frequent observation in a data set. For example, if total scores of a football tournament in every match were tabulated as 2, 4, 6, 5, 2, 4, and 2, then the mode of these observations is 2 scores
Bottling Company The mean, median and standard deviation are as follows: Bottle No. Oz mean median std dev The 95% confidence interval can be calculated as follows: CI = x +/- t* (s/?n) CI = 14.87 +/- 2.045 * (0.541 / ?30) CI = 14.87 +/- CI = 14.668 to 15.07 The null hypothesis is that the bottles are within the 95% confidence interval of 16oz as required by law. So the number of samples that fall within 15.8 to