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11) An industrial supplier has shipped truckload of teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of thes...

Question

11) An industrial supplier has shipped truckload of teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of these cartridges is excess of the 11 ounces printed on each cartridge: To check this claim sample of =19 cartridges are randomly selected from the shipment and carefully weighed. Summary statistics for the sample are: I2 ounces 19 ounce To determine whether the supplier'$ claim is true_ consider the test, Ho:u = Hvs Ha: 4 > IL; w

11) An industrial supplier has shipped truckload of teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of these cartridges is excess of the 11 ounces printed on each cartridge: To check this claim sample of =19 cartridges are randomly selected from the shipment and carefully weighed. Summary statistics for the sample are: I2 ounces 19 ounce To determine whether the supplier'$ claim is true_ consider the test, Ho:u = Hvs Ha: 4 > IL; where H the true mean weight of the cartridges. Calculate the value of the test statistic_ A) 2.000 B) 1.053 588 D) 20.000



Answers

Coke Cans Assume that cans of Coke are filled so that the actual amounts have a mean of 12.00 oz and a standard deviation of 0.11 oz. a. Find the probability that a single can of Coke has at least 12.19 oz. b. The 36 cans of Coke in Data Set 26 "Cola Weights and Volumes" in Appendix B have a mean of 12.19 oz. Find the probability that 36 random cans of Coke have a mean of at least 12.19 oz. c. Given the result from part (b), is it reasonable to believe that the cans are actually filled with a mean equal to 12.00 oz? If the mean is not equal to 12.00 oz, are consumers being cheated?

Number 10. A factory is located close to a city high school. The manager claims that the plant smokestacks and an average of no more than £350 of pollution per day. Someone to write that down as the null hypothesis, so no more. We're just gonna say equal to 350 as an ap stats project. That class plans a one sided hypothesis test with a critical value of £375 suppose that a standard deviation in daily pollution poundage is known to be 100 and £50 and the true mean is £385. So we have the true mean to be 3 85 with the standard deviation of 1 50. If the sample size is 100 days, what is the probability that the class will mistakenly fail to reject the factory managers false claim. So the alternative hypothesis would naturally be greater than 3 50 but we have the true meaning of 3 85% innovation, 1 50 with a sample size of 100. What is the probability that the class will mistakenly fail to reject the factory managers false claims? So mistakenly failing to reject is a type to error here. So for this class, they are going to reject past 3 75 which means they are going to fail to reject if it's less than 3 75. So let's find the probability that see is less than this test at 3 75 minus the true mean, in this case at 3 85 divided by the standard deviation 1 50 divided by route 100. Alright. Doing few calculations here. We get the probability that Z is less than negative 0.67 We can look on our chart or we can look on the um to 83 or 84 calculator using the normal CDF command. So drawing the picture to kind of get a look of what we're I'm going to type in. We have this negative 0.67 We want to find the probability of being to the left of it. Well, the lower regions where you start shading, technically that's a negative infinity will write it as negative 999 Upper bound is where we stop shading, which is negative 67 We're on a Z. Um standard normal curve. So I mean is zero center deviation is one. We plug this in and we get 0.2 five and looking at all of our answer choices. The answer choice that represents that is option B.

The following is a solution to # seven. It's over vacuum cleaners in the mean, wait, the mean weights of different types of vacuums or styles of vacuums, and the null hypothesis. In the alternative, they're always essentially the same thing. The null is that the means are equal This time we have three types of means. One was uh like bagged upright, one was bagless, upright, and the other one was a top canister or something. So the three means are equal to each other as the null. And then the alternative is that at least one of them is different, at least one of them. So not all of the new one. Mewtwo and mew theory equal the second step of this hypothesis test is to find the F. Star. Now you can use a table. Um since I can't really show a table on the computer, I'm gonna program something in the calculator. You I'm not going to show you how to program this, you can just google it if you want, but I want to find the critical value. I'm gonna call it F. Star. But first off to get F star, we need the alpha value which is the one that represents this area over here. And then I also need the degrees of freedom for the numerator and for the denominator. Now there are three categories. So the degrees of freedom for the numerator is the number of categories which is 3 -1. So too is the degrees of freedom for the numerator. And then the degrees of freedom for the denominator is the total number of data values minus the number of categories. So there are 18 data values in total minus the three categories. So 15 is the degrees of freedom for the denominator. So now I'm going to turn to my calculator. I'm going to go to program and I created this program called inverse F. And it's gonna ask me for my area to the right and remember that was point to one. That's my alpha value, degrees of freedom for the numerator was too and then the degrees of freedom for the denominator was 15. And that gives my critical value of about 6.36. So again you can find that and you can find that in um ah A table if that makes more sense to you or you can just write a program it's fine. So that represents this number here. So 6.36. So this area here is about 1% of the total curve point 01 So now I find the F statistic and you can do that with the formula. I don't suggest it. Especially if you have this calculator, Figure two Stat. And then edit, You can enter your data values in here for the L. one and L. two. And then if you go to stat and then a rover tests and go all the way down to a nova analysis of variance and go second L. One comma second L two com a second L. Three. And that tells me that those three columns or my categories is my data values. And if I press enter that gives me essentially everything I need to know. So this F. Value that's my F statistic Which is about 12.1. So I go back here and it's about 12.1 which definitely lands in the rejection region. Um So that's my decision right? So reject h not because it's definitely in the rejection reason so I can reject h. Not another way to verify that is you can look at this p value. This p value is 7.4 times 10 to the negative fourth. So it's written in standard I'm not not standard form and certain in scientific notation. So I sometimes have students say Okay the P value is 7.4. It's bigger than the alpha. That's not true. It will never ever ever be bigger than one. Because it stands for probability it's a very very small number whenever you see it in scientific notation. So either way the P value is less than alpha which means we reject the null hypothesis. And then the fifth and final step of this is always to conclude. Okay what does that actually mean? It just means that there is sufficient evidence. There is sufficient evidence to suggest that at least one mean vacuum cleaner. Wait. I don't know how to spell vacuum. I think I have too many seeds but that's fine, vacuum cleaner wait. Yeah. Is different than the others. Okay. And that is the Unova test Unova hypothesis test for vacuum cleaners.

Let us read this question. A consumer group claims that the mean annual consumption of high fructose considered by a person in the United States is 48.8. So mean is 48 is 48.8. So this is going to be my It's not or my null hypothesis. What is my alternative hypothesis? It is going to be that this is not equal to 48 point it. Okay, a random sample of 1. 20. So let me just directly right the formula. My N is equal to 1. 20. So over here, I'm going to have root off 1 20 route off. 1. 20. Uh, a random sample of 1 20 people in the United States has mean off 49.5. So the mean is 49.5 14 9 0.5 minus. What does my hypothesized mean? 48.8 divided by what is the population standard deviation? It is 3.6. So this is 3.6. Let me use a calculator to find this. So this is 49.5 minus 48.8, divided by 3.6 multiplied by. Excuse me, sir. This is 0.194444 This is going to be multiplied by route off 1 20. So this is 2.13 myself. Statistic is coming out to be 2.13 What is my Alfa level? My Alfa level is 0.0 fight. So this is a two tailed test, my al 50.5 which means in a single tail my area should be 0.25 So if I find a critical value for 0.25 my critical values 1.96 So which means this is minus 1.96 and this is plus 1.96 And since my Z value is beyond 1.96 it falls somewhere in the rejection region. I will reject final hypothesis, So I will reject h not Okay, So I will reject. It's not so what can I say? The consumer group claims that the mean annual consumption is 48 pointed. So I will say that I have in a statistical evidence to say that the claim off the consumer group is false and this is going to be my answer

Size 14. We are considering a scenario where an airline wishes to estimate the weight of the paint on a fully painted aircraft of the type it flies. And we are given that in a sample of four re paintings, meaning the sample sizes for The average weight of the paint applied was £239. Therefore the sample mean is £239. Okay? With sample standard division of £8. So the value of s there is £8. Yeah. And we also told to assume that the weights of paint on aircraft are normally distributed. And with that information we shall be constructing in 99 .8% confidence interval for the population. So let's begin by stating the formula to be used here. Uh It's going to be expo plus or minus the critical value of T for the given level of significance. Uh multiplied by S whereby the spirit of and this is because the population sample uh this population standard deviation is not known now because we are dealing with the 99 28% confidence interval, the level of significance, alpha will be one minus 0.998, which equals 0.002. And when we have that, we're going to get 0.001. No, since this is a T distribution, we need to determine the degrees of freedom By N -1. And in this case we have 4 -1, which he calls three, We're now ready to get the critical value of T at the 0.001 level of significance and 3° of freedom that equals yeah, like 10 .215. Now, we're ready to substitute all we have into the equation into the formula. So you're going to get 239 class or minus 10.215 times eight, divided by the square root of four, so that equals 239 plus or minus 40 0.86. So the margin of error is 40.86. And in conclusion, the 99.8% confidence interval for the mean weight of paint on all such air. Yes, Crafts is given there in pounds


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