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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 4% of 70%. 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. anonymous
    • one year ago
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    @maha2458 @batman19991

  2. anonymous
    • one year ago
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    I don't know how to even start this D:

  3. batman19991
    • one year ago
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    the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%. 90% is 1.645, 95% is 1.96, 99% is 2.575 so 9 / 10 means 90 / 100 the Confidence Interval is 90% accurate, i dont know if this helps though

  4. anonymous
    • one year ago
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    ohhhh okay! yes of course it does! because now I just need the margin of error and I'm done!!! thank you so much :)

  5. batman19991
    • one year ago
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    "Margin of Error" is 6% and your welcome

  6. anonymous
    • one year ago
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    oh okay! lol you mean "4%" ? my problem says "4% of 70%". I have no idea what the "of 70%' part means..what do you think it could mean? of 70% of the students?

  7. batman19991
    • one year ago
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    idk sorry

  8. jim_thompson5910
    • one year ago
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    it means that 70% is the estimate of the true population proportion the 4% is the margin of error which allows some wiggle room. The true population proportion is probably not exactly 70%, but somewhere in the range from 66% to 74% (notice how I added and subtracted 4% from 70%)

  9. anonymous
    • one year ago
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    OHHHHH

  10. anonymous
    • one year ago
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    so in the formula\[1.96 * \sqrt{\frac{ p(1 - p) }{ n }}\] p would be "0.7", right?

  11. anonymous
    • one year ago
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    in my class they taught me "p" is proportion and "n" is sample size

  12. anonymous
    • one year ago
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    oh wait it would be "1.645" instead of "1.96" I think, because it's a 90% confidence interval, right?

  13. jim_thompson5910
    • one year ago
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    we weren't given the sample size though

  14. anonymous
    • one year ago
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    yeah that's what I found confusing :/ but the question only asks for the confidence interval, the margin of error, and the confidence interval....hm I wonder why it asks for the confidence interval twice ._.

  15. batman19991
    • one year ago
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    oh well sorry dude

  16. anonymous
    • one year ago
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    maybe the first time is the percentage "90%" and the second time is the "66% - 74%' range that you calculated?

  17. jim_thompson5910
    • one year ago
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    or you can say (0.66, 0.74)

  18. anonymous
    • one year ago
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    How is this? The confidence interval is 90% because the exam claims that nine times out of ten, the seniors will have an average score within 4% of 70%. The margin of error is 4%. The proportion would be 70%, meaning that the confidence interval is between 66% and 74%. In terms of the situation, this means that 66% - 77% of the seniors will have an average score, 90% of the time.

  19. jim_thompson5910
    • one year ago
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    The sample proportion would be 70% The population proportion is what we're after. It is unknown, but very likely to be somewhere between 66% and 74% (with 90% confidence)

  20. anonymous
    • one year ago
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    Oh okay! That sounds great!! Thank you so much!

  21. jim_thompson5910
    • one year ago
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    "this means that 66% - 77% of the seniors" that's an incorrect way to state it

  22. anonymous
    • one year ago
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    Sorry! I meant 66% - 74%. It was a typo lol

  23. jim_thompson5910
    • one year ago
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    the proportions we're dealing with aren't proportions of the population size they are test scores

  24. anonymous
    • one year ago
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    Oh so should I write "In terms of the situation, this means that 66% - 74% of the time, the test scores will be average, with 90% confidence."

  25. jim_thompson5910
    • one year ago
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    that's also incorrect

  26. anonymous
    • one year ago
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    would it be 66% - 74% of the time, the seniors will have an average test score?

  27. anonymous
    • one year ago
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    wait so the population proportion is somewhere between "66% - 74%"

  28. jim_thompson5910
    • one year ago
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    Here's what is going on The teacher gathered up a sample of students. Say n = 100. We weren't given this value of n, but let's say it's 100. The teacher tested the students and then found the sample mean to be 0.70 This sample mean is the average test score of just the sample. The actual population mean is unknown The teacher claims that the true population mean is somewhere between 0.66 and 0.74, again those two numbers are test scores. So the true population mean of the test scores could be 0.71 or 0.68. We don't know for sure since we're not 100% confident. The 90% confidence means there is some room for error and the true population mean could be outside of this range

  29. anonymous
    • one year ago
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    Ohhhh!!! I understand now. I think I was getting confused with "population mean" and "sample mean". Thank you so much. You truly are a life saver.

  30. jim_thompson5910
    • one year ago
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    you're welcome

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