Question: Suppose you administer a certain aptitude test to a random sample of 9 students in your high school, and the average score is 105. We want to determine the mean μ of the population of all students in the school. Assume a standard deviation of σ = 15 for the test.
1.What is the margin of error?
2.What would be the interval for a 98% confidence interval?
3.Write a sentence explaining the 98% confidence interval in context of the question.
4.Write a few sentences explaining what would happen if you wanted a 80% confidence interval?
5.Write a few sentences explaining what would happen if you changed the sample size to be 81 and kept a 98% confidence interval.
Answer: 1. margin of error,e= σ/sqrt(n)* z_c [z_c= critical z at c% confidence interval.]
=15/sqrt(9)*2.33 [z_c=2.33 at 98% CF]
2. Confidence interval=(x bar -e, x bar +e)=(105-11.65, 105+11.65)=(93.25, 116.65)(Ans.)
3. Middle 98% values lie in between 93.25 to 116.65.(Ans.)
4. e= (15/3) *1.29=6.45
Confidence interval=(x bar -e, x bar +e)=(105-6.45, 105+6.45)=(98.55, 111.45)(Ans.)
5. e= (15/sqrt(81))*2.33=3.88
Confidence interval=(x bar -e, x bar +e)=(105-3.88, 105+3.88)=(101.12, 108.88)(Ans.) Edit
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