Nt1330 Unit 2 Assignment 2 Agression Analysis

Based on the given sample of student test scores of 50, 60, 74, 83, 83, 90, 90, 92, and 95 after rearranging them from least to greatest. As the mean is based on the average of sum, the average of this sample is 79.67 or 80. The mode refers to numbers that appear the most in a sequence and in this case 83 and 90 both appear twice. Range calculates the difference between the largest and smallest number, which are 95 and 50 which have a difference of 45. The variance is the difference between the sum of squares divided by the sample size, which is the number in the sample minus one (Hansen & Myers, 2012), meaning it takes each number of the set and subtracts

With the 95% Confidence Interval for Mean, Median, and St Dev are as described above.

Theoretically from the recorded data the calculated mean, median, and mode will be the most accurate representation of the real world value. The difference between the highest recorded value and lowest recorded value is the range in the set of data. Standard deviation (s) is a quantity calculated to indicate an extend of deviation for a group of data as a whole (Marshall). This is calculated using:

Bigger sample size will give a narrower confidence interval range (more specific) outliers affect the mean but not the median – this is why the median is preferred here.mean

By using the box and whisker plot below, four outliers were calculated for points in the data set. Outliers were larger than 81 points, and smaller than 33 points.

normally distributed, the 95% of all values will be within 2 standard deviations from the mean.

2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.

Now that the data are sorted, one result is the range of the data from 0 to 20, i.e. 21. The median is the middle observation in an odd number of sorted data and halfway between the two center-most points

Based on the chart, the mean was calculated by adding up the sum of the list and divide 18, which the number of the total listed prices. The mean is 135,000, which mean the average of the listed price. Secondly, the median was calculated by listing the number in numerical order from lowest to highest and located the number in the middle 126,000. The median represents the middle number of the listed price. After calculating the median I located the minimum and maximum based the lowest and highest data, which are 48,000 and 338,000. These represent the range of the listed price. Lastly, I used the formula to get the

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:

Today, the discussion of what constitutes an outlier has been very understanding. You will be giving a specific measurement that will give you an objective standard of what constitutes an outlier. Determining outliers is very simple, you multiplying the interquartile range which is short for (IQR) by 1.5 will give us a way to determine whether a certain value is an outlier. For an example, if you subtract 1.5 x IQR from the first quartile, any data values

In order to observe whether there exists a data point that particularly differs from other data, we use boxplot to analyze the outliers:

|Mean |51.84444 | |3 |0 1 3 3 4 5 5 7 8 8 8 8 9 |

Table 2 shows the average for each sample with one outlier removed. The standard deviation is based on one removed outlier.

An average number typically suggests the ‘mean’ value in a data-set; however, in this analyse has been used two different types of averages which are: the mean and median. As the mean is considered to be the one giving the most accurate information about the average number of takings – since it’s about adding up all the numbers form the data-set and divided them by the total - while the median represents only the value that stands in the middle of the data-set; in that way resulting a common mistake such as: leaving one of the averages aside and not analysing the accurate information from the data-set. However, there are no evidences to prove that both numbers are correct, as both are considered to be types of averages in statistics.