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Manufacture Interesie output voltage of powe supPIT usec p = volts against H " 4 voltsPC. Output voltage assumednormally distributed with standard deviatiorvol...

Question

Manufacture Interesie output voltage of powe supPIT usec p = volts against H " 4 voltsPC. Output voltage assumednormally distributed with standard deviatiorvoltmanufacurer WishestestFind the boundary of the critical region for the Type error probability and sample size provided. Rouna Your answers= Iwo decimal places (€.9 98.76).0.01 and<3 <(b) & = 0.0S and<}<(c) & = 0.01 and<R<(d) a = 0.05 and<X<

manufacture Interesie output voltage of powe supPIT usec p = volts against H " 4 volts PC. Output voltage assumed normally distributed with standard deviatior volt manufacurer Wishes test Find the boundary of the critical region for the Type error probability and sample size provided. Rouna Your answers= Iwo decimal places (€.9 98.76). 0.01 and <3 < (b) & = 0.0S and <}< (c) & = 0.01 and <R< (d) a = 0.05 and <X<



Answers

A manufacturer is interested in the output voltage of a power supply used in a $\mathrm{PC}$. Output voltage is assumed to be normally distributed with standard deviation 0.25 volt, and the manufacturer wishes to test $H_{0}: \mu=5$ volts against $H_{1}: \mu \neq 5$ volts, using $n=8$ units. (a) The acceptance region is $4.85 \leq \bar{x} \leq 5.15 .$ Find the value of $\alpha$. (b) Find the power of the test for detecting a true mean output voltage of 5.1 volts.

After the voltage is normally distributed with a standard deviation of 0.25 votes and the manufacturer which is the test. That new hypothesis, it's not musicals five votes against Alternative view is not equal five. Using an equal Z. This question is challenging our understanding of the language hypothesis test and how to properly implement a hypothesis test in a We want to answer the acceptance region is X between 4.85 and 5.15. What is after refers to the significance level that were to reject the hypothesis and songs from the probability of making the same point error. It is equal to the area in the tales outside of 4.85 and 5.5. So to reduce this area, we first need to find the corresponding to these critical boundary regions. Zero his excellency what we're seeing very and 4.85 91.7. Thus also over two or the area. The left tail is easy lessons not equal 20.446. That's such a probability to take one error is alpha equals 10.892. Because the two tailed tests, we must point out for over two by two. That's original answer. Over to 1.446. But you need to be what is the power of the test? So the power is 19. Data where betas probability we can take to error for truth. The output of 5.1 bolted we have probability that x files between 4.85 and 5.15 given U equals 5.1. It was probably easier between these two, these fours. That is probabilities between negative 22.57, which is 0.7133. That's the power is one minus and value or 10.2867.

As the postage is normally distributed with a standard deviation of 0.25 votes. And the manufacturer, which is to test the null hypothesis, a charge that you conspired votes against. The alternative view does not equal five. Using sample size that we want to find the battery of the critical region. If the type one area probability is alpha given below with sample size, that most question is challenging our understanding of how to utilize and interpret a critical region, which determines whether or not we accept to reject the null hypothesis and about in a hypothesis test. So first reading the alpha equals 0.1, N equals eight. We need to determine what the sample size and is critical from the normal distribution is. Such that P is the lesson that gives, you know, equals P is greater than zero equals out for over two there's a two tailed tests continued is not equal five. Thus we put our alpha into two tails. Thus this gives, you know plus minus 2.56 in the normal distribution. Thus, from the given our Z equals x minus people receive editor and the X values are critical. Boundary regions are X equals plus or minus the signal over and new, which is 4.775 point +23. We proceed an identical steps through the through G to solve so for b alpha equals 230.1 and equal 16 0. This post minus 2.56. Now X is 4.84 to 5.16 as our boundary region values. And see for an equal eight alpha equals 80.50 is now plus or minus 1.96. The region values are 4.83 and 5.17, finally in D 1.96, again for Z, thus 4.75 point +13 and are critical boundary values.

As the postage is normally distributed with a standard deviation of 0.25 votes. And the manufacturer, which is to test the null hypothesis, a charge that you conspired votes against. The alternative view does not equal five. Using sample size that we want to find the battery of the critical region. If the type one area probability is alpha given below with sample size, that most question is challenging our understanding of how to utilize and interpret a critical region, which determines whether or not we accept to reject the null hypothesis and about in a hypothesis test. So first reading the alpha equals 0.1, N equals eight. We need to determine what the sample size and is critical from the normal distribution is. Such that P is the lesson that gives, you know, equals P is greater than zero equals out for over two there's a two tailed tests continued is not equal five. Thus we put our alpha into two tails. Thus this gives, you know plus minus 2.56 in the normal distribution. Thus, from the given our Z equals x minus people receive editor and the X values are critical. Boundary regions are X equals plus or minus the signal over and new, which is 4.775 point +23. We proceed an identical steps through the through G to solve so for b alpha equals 230.1 and equal 16 0. This post minus 2.56. Now X is 4.84 to 5.16 as our boundary region values. And see for an equal eight alpha equals 80.50 is now plus or minus 1.96. The region values are 4.83 and 5.17, finally in D 1.96, again for Z, thus 4.75 point +13 and are critical boundary values.

For this exercise, we are told that the true voltage is 250 million volts and then that there is a measurement error that is continuous and uniformly distributed From -3 to plus three million volts. The error adds on to the true voltage. And when it is measured, when the voltage is measured, it is rounded to the nearest milli volts so that it is discreet. So they measured voltage will be discreet and it will be uniformly distributed on the range On the interval 247- 253. This corresponds to the true voltage 250 Plus or -3 for part A Lier asked to provide the probability mass function for the measured voltage. Now it can be measured Anywhere from any of the integers from 247 Up to 253. So that is seven possible valleys. And since it is uniformly distributed, Each has an equal probability of occurring so each is 1/7. So this is our probability mass function. And for B were asked for the mean and the variance of the measured voltage so similar to the case for the continuous random variable, the mean for the discrete uniform random variable is also a plus B over to where a corresponds to the bottom of the interval and be the top. This gives us 250. And the variance is different for the discrete than a continuous, it is b minus a plus one squared minus one over 12. And this gives us a variance of four


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