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 that is 800. And then the total sum will be divided by the total number of test scores that 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 above equation
=
= 80
The median in the set of a given number is the number which appears in the center of the given set of scores after arranging them in ascending order. The median test score of the given set of test scores is 83. It is to calculate median for the data set that is having an odd number of scores, as in that case the median is the central number. But in the case of given test scores, there were even numbers, so we have two scores in the middle of given data. So the median, in this case, would be the mean of both central numbers. And 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 are ten that is even and as there is no middle number so 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 give test score. As in the given case, 83 is the most recurrent number, so this is our mode. An important thing that is necessary to be remembered about finding the mode of a data set of numbers is 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.
Range is calculated by calculating the difference between highest and lowest value.
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 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 standard deviation, the closer will be the data to the mean. Whereas the large value of SD shows, 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
Method of mean and median is the best for this data set as the data set was simple and there was high dispersion. When there is high dispersion then mean and median are the best representative.
Part 3
Step 1
Ho: 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
Population mean is 70
Sample mean is 80
Standard deviation is 13.90
Step 4
t = where Sx =
Sx= = 4.402
t = = = 2.27
Step 5
Reject Ho if tobt is different from tcrit.
The critical value for two-tail one sample t-test with d.f = 9 and α = .05 is 2.26
As the values are different so sample mean does not represent population mean.
Part 4
This information is useful to understand the descriptives of a given sample and to further apply inferential statistics on it.
