Data Analysis

Test Scores = 95, 92, 90, 90, 83, 83, 83, 74, 60, 50

Mean = 80

Median = 83

Mode = 83

Range = 45

Variance = 193

Standard deviation = 13.90

Calculations

Summary

To calculate the mean of given students’ test scores, we have to compute the sum of given test scores, which is 800. Then, the total sum will be divided by the total number of test scores, which is 10 in the given data set.

Mean =

Step 1

= 95 + 92 + 90 + 90 + 83 + 83 + 83 + 74 + 60 + 50

= 800

Step 2

N = 10

Step 3

Applying the Values to the above equation

=

= 80

The median in the set of a given number is the number that appears in the centre of the given set of scores after being arranged in ascending order. The median test score of the given set of test scores is 83. It is to calculate the median for the data set that has an odd number of scores, as in that case, the median is the central number. However, in the case of given test scores, there were even numbers, so we have two scores in the middle of the given data. So, the median, in this case, would be the mean of both central numbers. In this set, the median is 83.

Step 1

First of all, they will arranged in ascending order.

= 50, 60, 74, 83, 83, 83, 90, 90, 92, 95

The total number of items is ten, which is even, and as there is no middle number, the median will be the average of the middle two numbers within the given list. The middle two numbers are 83 and 83. So,

= (83 + 83) ÷ 2

= 166 ÷ 2

= 83

So, the median is 83

The mode of a data set of numbers is the number that occurs most frequently in the given test score. As in the given case, 83 is the most recurrent number, so this is our mode. An important thing that must be remembered when finding the mode of a data set of numbers is that there may not be any mode at all. If no number is repeating at all, then there is no mode. In contrast, a data set can have multiple modes when there are the same number of duplicate values.

The mode is the most repetitive number in the list, so in the given test scores

95, 92, 90, 90, 83, 83, 83, 74, 60, 50

Mode is 83.

The range is calculated by calculating the difference between the highest and lowest values.

So here 95, 92, 90, 90, 83, 83, 83, 74, 60, 50

The highest value is 95, and the lowest value is 50

= 95 – 50

= 45

So, the range of the given test score is 45.

Standard deviation (SD) is a measurement that shows how spread out the numbers are of any given set of data. The smaller the value of the standard deviation, the closer the data will be to the mean. Whereas the large value of SD shows that the data is spread out away from the mean of data. SD is represented by a Greek letter σ is calculated from the value of variance. Variance is the average of the squared difference of each score from the mean. As in the given data, the value of SD is 14, and that of variance is 193.2.

First of all, Variance will be calculated

Step 1

95 + 92 + 90 + 90 + 83 + 83 + 83 + 74 + 60 + 50 = 800

Divide this value by the total number of tests (10 scores)

800 ÷ 10 = 80

The mean score on the test was a score of 80.

Step 2,

We need to subtract the mean from each test score and square each result.

(95 – 80)² = (15)² = 225

(92 – 80)² = (12)² =144

(90 – 80)² = (10)² = 100

(90 – 80)² = (10)² = 100
(83 – 80)² = (3)² = 9
(83 – 80)² = (3)² = 9
(83 – 80)² = (3)² = 9
(74 – 80)² = (-6)² = 36
(60 – 80)² = (-20)² = 400

(50 – 80)² = (-30)² = 900

Step 3

Find the mean of these values. Add them all together:

225 + 144 + 100 + 100 + 9 + 9 + 9 + 36 + 400 + 900 = 1932

Divide this value by the total number of scores (10 scores)

1932 ÷ 10 = 193.2 (rounded to the nearest whole score)

The variance of the test scores is 193. The standard deviation is simply the square root of the variance.

σ = √193.2 = 13.90 (round to nearest whole test score = 14)

Part 2

The mean and median methods are the best for this data set as they are simple and have high dispersion. When there is high dispersion, then the mean and median are the best representative.

Part 3

Step 1

H_o: µ=70

H₁: µ ≠ 70

Step 2

Set the criterion.

α = 0.05

d.f = n-1= 10-1=9

Critical value according to the table = 2.26

Step 3

From the given data, we know that

The population mean is 70

The sample mean is 80

The standard deviation is 13.90

Step 4

t = where S_x =

S_x= = 4.402

t = = = 2.27

Step 5

Reject Ho if t_obt is different from t_crit.

The critical value for a two-tail one sample t-test with d.f = 9 and α = .05 is 2.26

As the values are different, the sample mean does not represent the population mean.

Part 4

This information is useful for understanding the descriptives of a given sample and applying inferential statistics to it further.