## 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

1. anonymous

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

(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

@CBARREDO1 what concepts/terms should I know to solve this problem?

4. anonymous

@kropot72 hey kropot, any ideas on how to solve these problems? Thank you

5. kropot72

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

@kropot72 Thank you, I got A done. Any ideas on B and C?

7. kropot72

What is your result for A, the 99% confidence interval?

8. anonymous

(.7375, .8425) lower and upper bound

9. anonymous

@kropot72

10. kropot72

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

The answer should be yes since .95 isn't within the interval of .7375 and .8425. Is that correct?

12. kropot72

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

Thank you kropot. So for part C, how do I go about solving that equation?

14. kropot72

The first step in solving is to square both sides. Next you need to rearrange the result to give n > expression.

15. anonymous

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

Also, I will need to go for a bit and come back later. Thanks again for your help!

17. kropot72

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

18. anonymous

@kro@kropot72 is this the equation I use? https://www.ltcconline.net/greenl/images/NumInts/img8.gif

19. anonymous

@kropot72

20. kropot72

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

N will be the result of that equation you posted?

22. anonymous

I got 350.221

23. kropot72

Yes. So rounding up, the sample size needs to be equal to or greater than 351.

24. anonymous

AH, I see. Thanks so much! Any other work I need to show besides what you posted?

25. kropot72

You're welcome. Hopefully what I have posted is enough :)

26. anonymous

Thank you!