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 5% of 75%. 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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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 5% of 75%. 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.

Mathematics
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Ans: 2*ME = 80%-6% = 72% ME = 36% ----------- ------- Calculate the confidence interval and explain what it means in terms of the situation. 90% CI:: 0.06 < p < 0.80 We have 90% confidence the population proportion is between 6% and 80%
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Where did you get 0.06% and 0.80% from?
What does "nine times out of ten" tell you?
That is the confidence interval right? 90%
good
What are the critical values for a 90% confidence interval?
The critical value is 1.645
Correct, so Zα/2=Z.05=1.645. (Don't worry about the notation if you haven't seen it before.) The margin of error is given to be 5%, or 0.05, since an average score is expected to fall within 5% of the estimated mean. The estimated mean itself is?
it should be a very small percentage
it should be a very small percentage
Thats where I get stuck, so the margin of error is 0.05 which answers part of the problem, but what do you mean by the estimated mean itself?
is there another formula?
5% is correct
Okay, all i really need is the final part,Calculate the confidence interval and explain what it means in terms of the situation.
If you can help me solve that, I'll be good
(x¯−Zα/2σ2n−−−√, x¯+Zα/2σ2n−−−√) x¯ is the estimated average score. According to the given info, that would be 75%, or 0.75. The exam creator says the scores will fall within 5% of a 75% score, so the margin of error is 5%, or 0.05. In the formula, σ2n−−−√ is the margin of error. The critical value for a 90% confidence interval is 1.645, so Zα/2=1.645. So to reiterate, the confidence interval would be (0.75−1.645×0.05, 0.75+1.645×0.05) Simplify.
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Can you help me simplify? would it be ( -0.04475 , 0.83225)? and what does it tell us?
Those are what i got
Your answer would be the interval, yes. I suggest you study up on the topic. It doesn't seem like you're really grasping the procedure here.
Can you tell me what the interval means?
a space between things, points, limits, etc.: an interval of ten. 4. Math. a. the totality of points on a line between two designated points or endpoints that may or may not be included. b. any generalization of this to higher dimensions, as a rectangle with sides parallel to the coordinate axes
alright thanks, ill give you the best response even though you just copied and pasted the same exact thing, and even copied the wrong problem lmao
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.
A little pointer! Good luck!!

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