MEDAL AND FAN! A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standard deviation were calculated. Based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 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.
this is all I have... is it correct?? Since the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%, that means there is a 90% confidence interval. 9 / 10 = .90 = 90% 90% = 1.645 1.645 * .9 = 1.4805 Margin of Error - 1.4805
@HELP!!!! @amistre64
You are correct, there is a 90% confidence interval. The margin of error is the distance from x-bar, the center of the confidence interval ,to the end of the interval. This distance is 6% of 80% which is given by: \[\large \frac{80\times6}{100}=4.8%\] Therefore the margin of error is 4.8% and the confidence interval is: ({80 - 4.8}%, {80 + 4.8}%)
so I keep it in percents??
The exam result is given as a percentage, so I believe the answers should also be percentages.
ok, thanks!
You're welcome :)
@kropot72 what does the confidence level mean in the situation presented?
Note that the question asks for the meaning of the confidence INTERVAL ( not the confidence level ) in the situation presented. A confidence interval gives the likely range for a population parameter based on sample information. This interval is formed using the standard normal probability distribution, assuming that the sample size was 30 or greater. In the given situation we can say that a 90% confidence interval for the average score of the population is (75.2%, 84.8%) and 90% of such intervals will contain the average score of the population.
Got it! Thank you so much! @kropot72
You're welcome :)
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