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anonymous
 one year ago
The manufacture of the ColorSmart5000 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 colorsmart5000 television set for five years. Of these 400 consumers, 316 say their colorsmart5000 television sets did not need a repair, whereas 84 say their colorsmart5000 television sets did need atleast one repair
anonymous
 one year ago
The manufacture of the ColorSmart5000 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 colorsmart5000 television set for five years. Of these 400 consumers, 316 say their colorsmart5000 television sets did not need a repair, whereas 84 say their colorsmart5000 television sets did need atleast one repair

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a. Find a 99 percent confident interval for the proportion of all colorsmart5000 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 colorsmart5000 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 phat, the sample proportion of colorsmart5000 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

CBARREDO1
 one year ago
Best ResponseYou've already chosen the best response.0(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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@CBARREDO1 what concepts/terms should I know to solve this problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@kropot72 hey kropot, any ideas on how to solve these problems? Thank you

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1An approximate 99% confidence interval for the population proportion p is given by \[\large \hat{p}\pm2.576\sqrt{\frac{\hat{p}(1\hat{p})}{n}}\] \[\large \hat{p}=\frac{316}{400}=\frac{79}{100}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@kropot72 Thank you, I got A done. Any ideas on B and C?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1What is your result for A, the 99% confidence interval?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(.7375, .8425) lower and upper bound

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1b. 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer should be yes since .95 isn't within the interval of .7375 and .8425. Is that correct?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Yes 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: \[\large 2.576\sqrt{\frac{0.95(10.95)}{n}}<0.03\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you kropot. So for part C, how do I go about solving that equation?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1The first step in solving is to square both sides. Next you need to rearrange the result to give n > expression.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, I will need to go for a bit and come back later. Thanks again for your help!

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1np. The question relates to confidence intervals for population proportions, with part c 'finding the size of samples (precision)'.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@kro@kropot72 is this the equation I use? https://www.ltcconline.net/greenl/images/NumInts/img8.gif

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Not really. When you work on the equation that I posted as I outlined to you, the result is: \[\large n>\frac{(2.576)^{2}\times0.0475}{(0.03)^{2}}=you\ can\ calculate\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0N will be the result of that equation you posted?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Yes. So rounding up, the sample size needs to be equal to or greater than 351.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AH, I see. Thanks so much! Any other work I need to show besides what you posted?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome. Hopefully what I have posted is enough :)
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