a field test for a new exam was given to randomly selected seniors. The exams were graded and the sample mean and standard deviation were calculated. Based on the results, the exam creator claims that on the same exam, 9 out of 10 times, seniors will have an average score within 6% to 80%. Is the confidence interval at 90%, 95% or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.

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Question

a field test for a new exam was given to randomly selected seniors. The exams were graded and the sample mean and standard deviation were calculated. Based on the results, the exam creator claims that on the same exam, 9 out of 10 times, seniors will have an average score within 6% to 80%. Is the confidence interval at 90%, 95% or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.

@ganeshie8 help

anonymous · Answer

@ganeshie8 !!!!

anonymous · Answer

@dan815 @thomas5267 @mathmale

anonymous · Answer

anyone????? pls ;(

anonymous · Answer

@kohai

anonymous · Answer

@misty1212

ganeshie8 · Answer

`9 out of 10 times` tells you the confidence level is 90%

anonymous · Answer

how do I find the margin of error??

anonymous · Answer

@ganeshie8

anonymous · Answer

Margin of error = Critical value x Standard deviation of the statistic

anonymous · Answer

OR Margin of error = Critical value x Standard error of the statistic

ganeshie8 · Answer

margin of error = 6% of 80% = 0.06*80 = 4.8%

ganeshie8 · Answer

then the confidence interval should be : 

$$\large (80-4.8,~~80+4.8)$$

$$\large (75.2,~~84.8)$$