The manufacturer of the ColorSmart-5000 television set claims that 95 percent of
ID: 3224747 • Letter: T
Question
The manufacturer of the ColorSmart-5000 television set claims that 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 406 consumers who have owned a ColorSmart-5000 television set for five years. Of these 406 consumers, 312 say that their ColorSmart-5000 television sets did not need repair, while 94 say that their ColorSmart-5000television sets did need at least one repair.
Letting p be the proportion of ColorSmart-5000 television sets that last five years without a single repair, set up the null and alternative hypotheses that the consumer group should use to attempt to show that the manufacturer’s claim is false.
Use critical values and the previously given sample information to test the hypotheses you set up in part a by setting equal to .10, .05, .01, and .001. How much evidence is there that the manufacturer’s claim is false?
The manufacturer of the ColorSmart-5000 television set claims that 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 406 consumers who have owned a ColorSmart-5000 television set for five years. Of these 406 consumers, 312 say that their ColorSmart-5000 television sets did not need repair, while 94 say that their ColorSmart-5000television sets did need at least one repair.
Explanation / Answer
Null Hypothesis : H0: The Manufactures' claim is true and proportions of TV need repairs in 5 year time is equal or greater than 0.95. p >= 0.95
Alternative Hypothesis : Ha : The Manufactures' claim is false and proportions of TV need repairs in 5 year time is less than 0.95. p < 0.95
First we calculate Test Statistic:
p = (312/406) = 0.7684
Z = ( 0.7684 - 0.95)/ sqrt ( 0.95 * 0.05 / 406) = 0.1816/ 0.0108 = -16.81
Critical values of Z
for alpha = 0.1 => Zcritical = -1.28
for alpha = 0.05 => Zcritical = -1.64
for alpha = 0.01 => Zcritical = -2.33
for alpha = 0.001 => Zcritical = -3.08
so we will reject the null hypothesis as all t values are less than t critical for left tailed tests.
And as P- value is 0 so we have extremely strong evidence that the manufacturer's claim is true
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