) A sample of an examination result of 200 students was made. It was found that 46 students had failed, 68 secured III class, 62 second class and the rest were placed in the first division. Are these figure commensurate with the general examination results which is in the ratio 2:3 : 3; 2 for various categories respectively. Test at 5% alpha Level.
H1: (H0): No significant difference. There is goodness of fit. The sample of 200 students’ result is commensurate with general exam results
H2: There is difference. There is no goodness of fit. The sample of 200 students’ result is commensurate with general exam
Observed(O)
Expected(E)
(O-E)2
46 40
68 60
62 60
24 40
Total
Answers
Step-by-step explanation:
unity unity on the properties of the math you have so plant
Answer:
The expected frequency for each class can be calculated as:
46 students failed, E=2002/8 = 40
68 students secured III class, E=2003/8 = 60
62 students secured II class, E=2003/8 = 60
24 students secured I class, E=2002/8 = 40
Step-by-step explanation:
The chi-square goodness-of-fit test is used to determine whether a sample data set matches a population. To perform this test, you need to calculate the chi-square statistic and the corresponding p-value.
To calculate the chi-square statistic, you need to use the following formula:
Where O is the observed frequency, E is the expected frequency, and the sum is over all the categories.
Given that the general examination results is in the ratio 2:3 : 3; 2 for various categories respectively.
The expected frequency for each class can be calculated as:
46 students failed, E=2002/8 = 40
68 students secured III class, E=2003/8 = 60
62 students secured II class, E=2003/8 = 60
24 students secured I class, E=2002/8 = 40
Using the above formula, we can calculate the chi-square statistic as:
χ2 = ((46-40)^2/40) + ((68-60)^2/60) + ((62-60)^2/60) + ((24-40)^2/40) = 2.5
To find the p-value, we need to use the chi-square distribution table, with the degrees of freedom equal to the number of categories - 1, which is 3.
With a chi-square statistic of 2.5 and 3 degrees of freedom, and a significance level of 0.05, the corresponding p-value from the chi-square distribution table is >0.05, which is greater than the significance level.
Since the p-value is greater than the significance level, we fail to reject the null hypothesis. We can conclude that there is no significant difference between the sample data and the population, and that the sample of 200 students' results is commensurate with the general examination results.
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