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anonymous

  • one year ago

The manufacture of the ColorSmart-5000 television set would like to claim 95 percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects 400 consumers who have owned a colorsmart-5000 television set for five years. Of these 400 consumers, 316 say their colorsmart-5000 television sets did not need a repair, whereas 84 say their colorsmart-5000 television sets did need atleast one repair

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  1. anonymous
    • one year ago
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    a. Find a 99 percent confident interval for the proportion of all colorsmart-5000 television sets that have lasted at least five years without needing a single repair b. Does this confidence interval provide strong evidence that the percentage of colorsmart-5000 television sets that last atleast five years without a single repair is less than the 95 percent claimed by the manufacturer? Yes or no? Expain. c. Determine the sample size needed in order to be 99 percent confident that p-hat, the sample proportion of colorsmart-5000 television sets that last atleast five years without a single repair, is within .03 of p, the true proportion of sets that last atleast five years without a single repair

  2. CBARREDO1
    • one year ago
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    (phat - p)/sqrt(phat*q/n) ∼ n(0, 1) P(0.79 - 2.575829*(sqrt(0.79*0.21/400)) < π < 0.79 + 2.575829*(sqrt(0.79*0.21/400)) = 0.99 P(0.79 - 0.05 < π < 0.79 + 0.05) = 0.99 P( 0.74 < π < 0.84) = 0.99 amplitude = 0.1 P(0.79 - 1.96*(sqrt(0.79*0.21/400)) < π < 0.79 + 1.96*(sqrt(0.79*0.21/400)) = 0.95 P(0.79 - 0.04 < π < 0.79 + 0.04) = 0.95 P(0.75 < π < 0.83) = 0.95 amplitude = 0.08 B) Ever, the interval at 99% is wider that the interval at 95%. The reason is the different z value used in both of them!

  3. anonymous
    • one year ago
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    @CBARREDO1 what concepts/terms should I know to solve this problem?

  4. anonymous
    • one year ago
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    @kropot72 hey kropot, any ideas on how to solve these problems? Thank you

  5. kropot72
    • one year ago
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    An approximate 99% confidence interval for the population proportion p is given by \[\large \hat{p}\pm2.576\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] \[\large \hat{p}=\frac{316}{400}=\frac{79}{100}\]

  6. anonymous
    • one year ago
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    @kropot72 Thank you, I got A done. Any ideas on B and C?

  7. kropot72
    • one year ago
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    What is your result for A, the 99% confidence interval?

  8. anonymous
    • one year ago
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    (.7375, .8425) lower and upper bound

  9. anonymous
    • one year ago
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    @kropot72

  10. kropot72
    • one year ago
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    b. A 99% confidence interval for the population mean is an interval such that on average, 99 out of every 100 such intervals will contain the population mean. The manufacturer claims that 95% of the sets last 5 years without needing repair, so effectively the manufacturer is claiming a population mean of 0.95. Compare this with your correct result for the 99% CI.

  11. anonymous
    • one year ago
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    The answer should be yes since .95 isn't within the interval of .7375 and .8425. Is that correct?

  12. kropot72
    • one year ago
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    Yes is correct for part b, together with an appropriate explanation. For part c you need to solve the following equation to find the value of n, the sample size: \[\large 2.576\sqrt{\frac{0.95(1-0.95)}{n}}<0.03\]

  13. anonymous
    • one year ago
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    Thank you kropot. So for part C, how do I go about solving that equation?

  14. kropot72
    • one year ago
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    The first step in solving is to square both sides. Next you need to rearrange the result to give n > expression.

  15. anonymous
    • one year ago
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    Ok, thank you. Is there a name for this type of problem/concept that I can look up to better help me understand how to solve this?

  16. anonymous
    • one year ago
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    Also, I will need to go for a bit and come back later. Thanks again for your help!

  17. kropot72
    • one year ago
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    np. The question relates to confidence intervals for population proportions, with part c 'finding the size of samples (precision)'.

  18. anonymous
    • one year ago
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    @kro@kropot72 is this the equation I use? https://www.ltcconline.net/greenl/images/NumInts/img8.gif

  19. anonymous
    • one year ago
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    @kropot72

  20. kropot72
    • one year ago
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    Not really. When you work on the equation that I posted as I outlined to you, the result is: \[\large n>\frac{(2.576)^{2}\times0.0475}{(0.03)^{2}}=you\ can\ calculate\]

  21. anonymous
    • one year ago
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    N will be the result of that equation you posted?

  22. anonymous
    • one year ago
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    I got 350.221

  23. kropot72
    • one year ago
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    Yes. So rounding up, the sample size needs to be equal to or greater than 351.

  24. anonymous
    • one year ago
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    AH, I see. Thanks so much! Any other work I need to show besides what you posted?

  25. kropot72
    • one year ago
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    You're welcome. Hopefully what I have posted is enough :)

  26. anonymous
    • one year ago
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    Thank you!

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