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anonymous

  • one year ago

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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  1. freshkeith56
    • one year ago
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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%

  2. freshkeith56
    • one year ago
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    Need more explanation?

  3. freshkeith56
    • one year ago
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    Or are you good?

  4. anonymous
    • one year ago
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    Where did you get 0.06% and 0.80% from?

  5. freshkeith56
    • one year ago
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    What does "nine times out of ten" tell you?

  6. anonymous
    • one year ago
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    That is the confidence interval right? 90%

  7. freshkeith56
    • one year ago
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    good

  8. freshkeith56
    • one year ago
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    What are the critical values for a 90% confidence interval?

  9. anonymous
    • one year ago
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    The critical value is 1.645

  10. freshkeith56
    • one year ago
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    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?

  11. freshkeith56
    • one year ago
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    it should be a very small percentage

  12. freshkeith56
    • one year ago
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    it should be a very small percentage

  13. anonymous
    • one year ago
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    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?

  14. anonymous
    • one year ago
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    is there another formula?

  15. freshkeith56
    • one year ago
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    5% is correct

  16. anonymous
    • one year ago
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    Okay, all i really need is the final part,Calculate the confidence interval and explain what it means in terms of the situation.

  17. anonymous
    • one year ago
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    If you can help me solve that, I'll be good

  18. freshkeith56
    • one year ago
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    (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.

  19. freshkeith56
    • one year ago
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  20. anonymous
    • one year ago
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    Can you help me simplify? would it be ( -0.04475 , 0.83225)? and what does it tell us?

  21. anonymous
    • one year ago
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    Those are what i got

  22. freshkeith56
    • one year ago
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    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.

  23. anonymous
    • one year ago
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    Can you tell me what the interval means?

  24. freshkeith56
    • one year ago
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    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

  25. anonymous
    • one year ago
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    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

  26. freshkeith56
    • one year ago
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    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.

  27. freshkeith56
    • one year ago
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    A little pointer! Good luck!!

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