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A computer manufacturer estimates that its line of minicomputershas, on average, 8.5 days of downtime per year. To testthis claim, a researcher contacts seven compa...

Question

A computer manufacturer estimates that its line of minicomputershas, on average, 8.5 days of downtime per year. To testthis claim, a researcher contacts seven companies that own one ofthese computers and is allowed to access company computer records.It is determined that, for the sample, the average number ofdowntime days is 5.8, with a sample standard deviationof 1.5 days. Assuming that number of downtime days isnormally distributed, test to determine whether these minicomputersactually average

A computer manufacturer estimates that its line of minicomputers has, on average, 8.5 days of downtime per year. To test this claim, a researcher contacts seven companies that own one of these computers and is allowed to access company computer records. It is determined that, for the sample, the average number of downtime days is 5.8, with a sample standard deviation of 1.5 days. Assuming that number of downtime days is normally distributed, test to determine whether these minicomputers actually average 8.5 days of downtime in the entire population. Let α = .01. Round your answer to 2 decimal places.)- A. The value of the test statistic is B. do we fail to reject the null hypothesis or reject the null hypothesis



Answers

The downtime per day for a computing facility has mean 4 hours and standard deviation. .8 hour. a. Suppose that we want to compute probabilities about the average daily downtime for a period of 30 days.i. What assumptions must be true to use the result of Theorem 7.4 to obtain a valid approximation for probabilities about the average daily downtime? i. Under the assumptions described in part (i), what is the approximate probability that the average daily downtime for a period of 30 days is between 1 and 5 hours? b. Under the assumptions described in part (a), what is the approximate probability that the total downtime for a period of 30 days is less than 115 hours?

Problem 26 which in a the null hypothesis it is that you is bigger than or equal 5.8. And Alternative five busted is that new is smaller than 5.8 toe Remind. The T Value T value is equal to explore minus m. Note. Over s over square root of X, which is 4.9 minus 5.8 over 1.8 over square root for 35 which is equal to negative 2.958 Question. The very value is the number are in the colon for table five containing the T value degree of freedom equal toe 35 minus one, which is 34. So the be is one of 10.45 For which, and see if the prevailing smaller than the significance level that the high positive rejected so be is smaller than a point or one. So we reject They're not high Protestants for goods in these. So if we can say that there is sufficient evidence for injecting likely

Let us read this question. A light bulb manufacturer guarantees that the mean life off a certain type of light bulb is at least 7 50 hours. So mean life is at least 7 50 years, which means it is greater than or equal toe 7 50 years. So the complement of this is that it is less than 7 50 hours. So let's say that this is a null hypothesis. The first one. The claim forms the null hypothesis and the compliment over here is the alternative hypothesis. A random sample of 25 light bulbs so our end is equal to 25 r n is equal to 25 and they're mean is 7 45. So x bar is equal to 7. 45. Standard deviation off the population is 60. Our sigma is equal to 60. So let's just put in the values. Misers to district will be equal toe 7 45 minus 7 50. This is what I'm checking it against. Upon Sigma is 60 upon root off. What is my end in his 25 which is nothing but five. So if I use a calculator to find this, this is going to be 7 45 minus 7 50 divided by 16 multiplied by five or minus 4.16 Now this is a This is using magnitude, right? Minus 4.16 This Z statistic is used in number. Okay, My Alfa is 0.2 So what is going to be my critical Z value? It is 2.6 My Zed star. Let me say that my said star is 2.6 and this is 2.6 and my value off minus 4.16 falls somewhere around here. So this is in the non rejection region, right? What is my alternative? My, my alternative. Excuse me. My alternative is less than so. This is going to be minus 2.6 My critical value is going to be minus 2.6 So it is not going to be here, but rather it is going to be on the negative side. So this is minus 2.6 My critical value, minus 2.6 on my value falls in the rejection region falls in this region in the rejection region. So I will reject my null hypothesis. What was my null hypothesis? My null hypothesis was the claim so that I can say that I have enough statistical evidence to say that the claim or the guarantee off the light bulb manufacturer is actually false. Okay, He said that it is at least 7. 50. It is at least 7 50 years, but that turns out to be false. That claim is false, and this is going to be my answer.

Were given information on samples of women who are insured and uninsured in terms of how long they spent in the hospital for routine childbirth. And we're interested, if at the significance level of Alfa equals 0.1 If the means of the two different groups of women the two populations of women insured and uninsured, are different. We've been asked to you test this using the P value approach and to also make a 99% confidence interval for the difference in means. So we want to know if there's any difference in the mean time, the mean length of stay. So the no hypothesis is that there is no difference. And the alternative hypothesis is then that there is a difference so we can go ahead and calculate the test statistic. We don't have the population standard deviations. We're only using the sample standard deviations. Therefore, our test statistic is distributed according to the student's T distribution, and it's given as follows. And so we have all of these values in the summary table provided here so we can go ahead and plug goes in and we get the following and this comes out to two point 385 Now we confined our associated P value with that test statistic value. And so there's two things. There is the test value itself, and the degrees of freedom is n minus one. So that's 15. So we can go ahead to a normal table, a rather tea table. We have 15 degrees of freedom and t of 2.385 is between these two values, which means that our key value we call that this is a a two tailed test. Our alternative hypothesis is that they're not equal. Therefore, RPI value is between 0.2 and 0.5 And since we're testing this at the Alfa equals 0.1 significance level, we can say that RPI value is bigger than Alfa and therefore we failed to reject the null hypothesis and we can conclude by saying that there is insufficient evidence to support the claim that the mean hospital time stays are different now. We've also been asked to make a 99% confidence interval, and this is of the form the meantime for insured, minus the mean time for uninsured teas of ALF over to times the standard air, which is the denominator in the test statistic. We can use the table to find tease of Alfa over to at 15 degrees of freedom and we have an area of 0.1 in two tails. So we have 2.947 So now we have all the values we need to solve this equation or this solve this interval and this gives us a confidence interval ranging from negative zero point 094 to 0.8 94 So we can say that we're 99% confident that the true difference between the two populations the insured women and the uninsured women is between negative 0.94 and 0.894 days.

Part in this question. The first step is to arrange data sending would. So if you do that looks something like this. And since our data is even real, you take average of two middle values. And since two US Jane over to this will give us 30 to 30 three, which is stage five, that's our me. Now, if to use the original data but in the values of the medians and read to give runs a B above below so you can see that V is the observed number fronts served Brundidge, which is for thank you. No, I need to find the test statistic. See, star. So now we know that now that Frances is that the downtime data is ready. Number done. Data it is. And alternate buses H A is that there is a trend in the downturn data. So it's not friend. Okay, find the mean under standard deviations. That statistic so is equal to 21 times two who for plus in two, that's what. Which is two times trolls over. So no approximation for mean is find normal approximation for the standard deviation. Spare two times two rooms to to to times and to minus one medicine. Okay, it's a very thing is for unpleasant screwed two squared two and one plus to minus one player thing, and you have to before. So that's the normal approximation for the standard deviation. So now we have a data we need to find the test statistic value. Z Star is equal to V star minus mu V over. I'm just plugging all the values we have for 13 over four, which is equal to 3.7 native to. So that is a test statistic. Value on now From this, when the minutes have put P values is your pins your to P is Is your your jury zero from So In part C begin concluded. P value is less sure. Please your five so we can reject. I didn't know my buses it This so there's sufficient evidence to indicate that there's a trend and don't time did. Therefore, data is not right


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