From the above test scores given, the mean is calculated by finding the average of the given values. This is achieved by adding all values, and then dividing by the total number of values denoted as “n.” In our case, n is equal to 10. It is calculated as (95+92+90+90+83+83+83+74+60+50), which is equal to 800 and then divided by 10. Therefore, the mean is equal to 80.
Median is a value which divides the data sample into two halves. This is determined by first ordering the values in the data set from the lowest number to the highest and selecting the middle value. It is easier to identify the median when “n” is an odd number. However, in a case where “n” is an even number, the median is calculated by finding the average of the two middle numbers. In our case, n is an even; therefore, the median is calculated as (83+83)/2. The median is 83.
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Mode, on the other hand, is the value that appears the highest number of times in a given data set. In our case, 83 appears most frequently compare to other values. Therefore, the mode is 83
The range indicates how much is the maximum value in a set of data is different from the minimum value in the same set of data. In this case, the range is 45. It is determined by subtracting the minimum value from the maximum value. Here, the maximum value is 95, and the lowest value is 50. The difference between the two numbers is 45.
From my calculations, the variance is 193.2. To arrive at this value, calculate the difference between each of the values and the mean in a given data set. Next, square the deviations and find the average of the squared distance from the mean as follows (95-80) 2 + (92-80) 2 + (90-80) 2 + (90-80) 2 + (83-80) 2 + (83-80) 2 + (83-80) 2 + (74-80) 2 + (60-80) 2 + (50-80) 2 = 1932. Dividing the value by 10, we get the variance and is denoted by S 2 . Standard deviation is computed by taking the positive square root of the variance (Sedkaoui, 2018) . Calculating the square root of 193.2, it gives a standard deviation of 13.8996 which can be rounded off to 13.9. S denotes standard.
One sample t-test is appropriate for this particular data set. It tests whether there is a difference between means of a continuous level data which is normally distributed (Peck, Short & Olsen, n.d.) . It compares the sample mean with a hypothetical mean, usually, the population means. One sample t-test is useful when we have one sample mean, and the population standard deviation is unknown. To conduct a one sample t-test, one should define the null and alternative hypothesis, state the alpha level of significance, calculate the degrees of freedom, state the decision rule, calculate the test statistic, report the results, and interpret the results as shown below;
H 0 : µ = 70
H 1 : µ ≠ 70
Significance level α = 0.05
Degrees of freedom, n-1 = 10-1 = 9
At α = 0.05 and 9 degrees of freedom, the t critical value is 2.262. The null hypothesis is rejected if the test statistic is less than -2.262 or is greater than 2.262.
Test statistic, t = (X̅- µ)/ (S/√n)
= (80-70) / (13.9/√10)
= 2.275
The test statistic is greater than 2.262. Therefore, the null hypothesis is rejected and concluded that the population means is significantly different from the sample mean.
This information is useful when there is a well-known and commonly established means, and a researcher wants to test against its difference. For instance, a situation when one wants to know if the mean number of patients attending hospital on daily in his/ her village is different from a defined population mean.
References
Peck, R., Short, T., & Olsen, C. Introduction to statistics and data analysis .
Sedkaoui, S. (2018). Data analytics and big data . London: ISTE Ltd / John Wiley & Sons, Inc.
