The marks of DMT students results in June 2020 sessional examinations were normally disyributed with a mean pass mark of 9 and a standard deviation pa

aflacatn 2021-03-05 Answered
The marks of DMT students results in June 2020 sessional examinations were normally disyributed with a mean pass mark of 9 and a standard deviation pass mark of 0.15. After moderation, a sample of 30 papers was selected to see if the mean pass mark had changed. The mean pass mark of the sample was 8.95. a) Find the 95% confidence interval of students mean mark. b) Calculate for the critical regions of the 95% confidence intervals. c) Using your results in "a" and "b" above, is there evidence of a change in the mean pass mark of the DMT students.
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Willie
Answered 2021-03-06 Author has 95 answers

Step 1 Let 'X' be the marks of students X N (9, 0.152) a) The formula for the confidence interval is C.I=[x ± zα2. σn] The sample size and sample mean are given to be, n=30
x=8.95 The z-critical value at α=0.05 is 1.96 C.I=[8.95± (1.96). 0.1530]
=[8.95± 0.0537]
=[8.95, 9.0037] b) The critical regions for 95% confidence interval are (1  0.952)= ± 0.025 Step 2 c) The hypotesis are: H0: μ=9
H1: μ9 The test statistics is given by: z= x  μσn
= 8.95  90.1530
= 0.050.0274
= 1.825 The p-value lies between 0.0336 and 0.0344 p-value is greater than the alpha level of significance 0.025, ther null hypotesis is accepted. Conclusion: There is a lack of evidence in the mean pass marks of the students.

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