The manufacture of the ColorSmart-5000 television set would like to claim 95 per
ID: 3333774 • Letter: T
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 at least one repair.
Problem 2: 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 at least one repair. a) What is the sample proportion? Answer Canvas Quiz 6-4 (1 point) b) 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. What is the critical value (Z-/2 or ton)? Answer Canvas Quiz 6-5. (1 point) What is the lower limit of the interval? Answer Canvas Quiz 6-6. (1 point) What is the upper limit of the interval? Answer Canvas Quiz 6-7. (1 point) c) Does this confidence interval provide strong evidence that the percentage of ColorSmart-5000 television sets that last at least five years without a single repair is less than the 95 percent claimed by the manufacturer? Circle your answer and explain. Use your answer to answer Canvas Quiz 6-8. (1 Point) Yes No Explain: d) Determine the sample size needed in order to be 99 percent confident that p, the sample proportion of ColorSmart-5000 television sets that last at least five years without a single repair, is within .03 of p, the true proportion of sets that last at least five years without a single repair. 1. Practical choice: Use your answer from part a) and answer Canvas Quiz 6-9. (1 point) 2. Conservative Choice: Use p = 0.5, and answer Canvas Quiz 6-10. (1 point)Explanation / Answer
a.
RADITIONAL METHOD
given that,
possibile chances (x)=316
sample size(n)=400
success rate ( p )= x/n = 0.79
I.
sample proportion = 0.79
standard error = Sqrt ( (0.79*0.21) /400) )
= 0.02
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.02
= 0.04
III.
CI = [ p ± margin of error ]
confidence interval = [0.79 ± 0.04]
= [ 0.75 , 0.83]
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DIRECT METHOD
given that,
possibile chances (x)=316
sample size(n)=400
success rate ( p )= x/n = 0.79
CI = confidence interval
confidence interval = [ 0.79 ± 1.96 * Sqrt ( (0.79*0.21) /400) ) ]
= [0.79 - 1.96 * Sqrt ( (0.79*0.21) /400) , 0.79 + 1.96 * Sqrt ( (0.79*0.21) /400) ]
= [0.75 , 0.83]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.75 , 0.83] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
b.
TRADITIONAL METHOD
given that,
possibile chances (x)=316
sample size(n)=400
success rate ( p )= x/n = 0.79
I.
sample proportion = 0.79
standard error = Sqrt ( (0.79*0.21) /400) )
= 0.02
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
margin of error = 2.326 * 0.02
= 0.047
III.
CI = [ p ± margin of error ]
confidence interval = [0.79 ± 0.047]
= [ 0.743 , 0.837]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=316
sample size(n)=400
success rate ( p )= x/n = 0.79
CI = confidence interval
confidence interval = [ 0.79 ± 2.326 * Sqrt ( (0.79*0.21) /400) ) ]
= [0.79 - 2.326 * Sqrt ( (0.79*0.21) /400) , 0.79 + 2.326 * Sqrt ( (0.79*0.21) /400) ]
= [0.743 , 0.837]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.743 , 0.837] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
c.
yes
d.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.79
ME = 0.047
n = ( 2.576 / 0.047 )^2 * 0.79*0.21
= 498.359 ~ 499
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