A random sample of 3586 students from various areas is taken to compare their performance in Biology, based on the final exam mark (low: below 40%, medium: between 40% and 60%, high: above 60%), in relation to the amount of private tutoring they receive. The results are summarized in the table below Performance in Biology Tutoring Low Medium High Total No tutoring 46 (11%) 168 (41%) 196 (48%) 410 (100%) Some tutoring 100(5%) 572 (31%) 1148 (63%) 1820 (100%) Frequent tutoring 32 (2%) 248 (18%) 1076 (79%) 1356 (100%) Total 178(5%) 988 (28%) 2420 (67%) 3586 (100%) Based on the data in the table, and without doing a significance test, how would you describe the relationship between receiving private tutoring and performance in Biology? Calculate the Chi-Square statistic and use it to test for independence, using a 10% significance level. What do you conclude? Would you conclude that private tutoring improves the performance in Biology? Briefly justify your
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
A random sample of 3586 students from various areas is taken to compare their performance in Biology, based on the final exam mark (low: below 40%, medium: between 40% and 60%, high: above 60%), in relation to the amount of private tutoring they receive. The results are summarized in the table below
|
|
Performance in |
Biology |
||||
|
Tutoring |
Low |
Medium |
High |
Total |
||
|
No tutoring |
46 (11%) |
168 |
(41%) |
196 (48%) |
410 (100%) |
|
|
Some tutoring |
100(5%) |
572 |
(31%) |
1148 (63%) |
1820 (100%) |
|
|
Frequent tutoring |
32 (2%) |
248 |
(18%) |
1076 (79%) |
1356 (100%) |
|
|
Total |
178(5%) |
988 |
(28%) |
2420 (67%) |
3586 (100%) |
|
- Based on the data in the table, and without doing a significance test, how would you describe the relationship between receiving private tutoring and performance in Biology?
- Calculate the Chi-Square statistic and use it to test for independence, using a 10% significance level. What do you conclude?
- Would you conclude that private tutoring improves the performance in Biology? Briefly justify your
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