Question

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

Answer 1

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

Answer 2

(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!

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

(.7375, .8425) lower and upper bound

Answer 9

@kropot72

Answer 10

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.

Answer 11

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

Answer 12

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:
\[\large 2.576\sqrt{\frac{0.95(1-0.95)}{n}}<0.03\]

Answer 13

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

Answer 14

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

Answer 15

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?

Answer 16

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

Answer 17

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

Answer 18

@kro@kropot72 is this the equation I use?

https://www.ltcconline.net/greenl/images/NumInts/img8.gif

Answer 19

@kropot72

Answer 20

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

Answer 21

N will be the result of that equation you posted?

Answer 22

I got 350.221

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

Thank you!