Question

In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures. 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, what can we conclude about the time between failures? Use the built-in statistical features of your calculator. State the hypotheses: Report the test statistic: Report the p-value: State your conclusion:

          In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures. 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, what can we conclude about the time between failures? Use the built-in statistical features of your calculator. State the hypotheses: Report the test statistic: Report the p-value: State your conclusion:
        
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In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures. 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, what can we conclude about the time between failures? Use the built-in statistical features of your calculator. State the hypotheses: Report the test statistic: Report the p-value: State your conclusion:

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures. 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, what can we conclude about the time between failures? Use the built-in statistical features of your calculator. State the hypotheses: Report the test statistic: Report the p-value: State your conclusion:
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In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures: 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours.

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In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 12 modified components resulted in the following times (in hours) between failures: 518, 548, 561, 523, 536, 522, 499, 538, 557, 528, 563, 530. At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours. Assume the data comes from a population that is essentially normal.

Madhur L.


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Transcript

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00:01 Hello students, according to the given question, the null hypothesis can be written as h0 such that u is equal to 520 and the alternative hypothesis h1 such that u is greater than 520.
00:17 So here the population mean, population mean that is u will be equal to 520 and the sample mean that is is x will be equal to 5307 .1 so by doing the given data as the sum of squares divided with total number of squares.
00:48 So here's the standard deviation that is sigma will be equal to 20 .7013.
00:57 Here number that is n is 10.
01:01 So the test statistic will be p will be equal to 10.
01:05 The value x minus u is divided by sigma by root n.
01:12 So by substituting we get which is equal to 537 .1 minus 520 is divided by 20 .7013 is divided by square root of 10.
01:29 So by solving it we get to 0 .612.
01:32 Therefore, t value will be 2 .612.
01:38 Now the critical value, that is t alpha with n minus 1 degrees of freedom, will be equal to 1 .833.
01:48 So we got t value, that is t statistic value, that is 2 .612, and the t alpha critical value will be 1 .833.
02:00 So, coming to the decision, hence the decision...
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