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Qunton 2Achocolate tactory crcates chocolate eggs with surprise toys inside_ Order Drevunt choking: the national health and safety ' board demands that toYs mu...

Question

Qunton 2Achocolate tactory crcates chocolate eggs with surprise toys inside_ Order Drevunt choking: the national health and safety ' board demands that toYs must be larger than inches wlde- order remain open, the factory must prove that the averge size of thelr toys strictly greater than inches wide. They take sample 0f 179 eggs and measures the toys inside; finding an averjge 0f 1.58 Inches with standard deviation of 0.4. Do thev have enough evidence kecp their factory open? Use 0.01 teve

Qunton 2 Achocolate tactory crcates chocolate eggs with surprise toys inside_ Order Drevunt choking: the national health and safety ' board demands that toYs must be larger than inches wlde- order remain open, the factory must prove that the averge size of thelr toys strictly greater than inches wide. They take sample 0f 179 eggs and measures the toys inside; finding an averjge 0f 1.58 Inches with standard deviation of 0.4. Do thev have enough evidence kecp their factory open? Use 0.01 tevel of signlfcance State the null and the alternative hypotheses. Select the appropriate inference procedure and verify conditions for using Specify the rejection region: C Calculate the value of the test statistic; D. State the test decision and conclusion;



Answers

(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value(s) and identify the rejection region$(s),(c)$ find the standardized test statistic $z,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. If convenient, use technology. A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 920 milligrams. A random sample of 44 breakfast sandwiches has a mean sodium content of 925 milligrams. Assume the population standard deviation is 18 milligrams. At $\alpha=0.10,$ do you have enough evidence to reject the restaurant's claim?

What's up, stat? Cats. My name is Aaron, and in this video, we're gonna be looking at an example performing an up test and using the traditional method of hypothesis testing to do so. So we're looking at Is there a difference in the variation of carb content between chocolate and non chocolate Candies? And we're gonna be using the output level 0.1 And our sample size for a chocolate is 13 and our sample size for non chocolate is 11. So the first thing we're gonna do is we're going to stay. Our hypotheses and identify are clean. So are no hypotheses is that the variation between the two types of candy is gonna be equal. So chocolate and non chocolate and our alternative hypothesis is that the variances of the two Candies are not gonna be so because they're not equal. This tells us that we're gonna do a two tailed test and our new out the level it's going to be 0.1 divided by two. That's going to give us 05 as our new off the level and our alternative hypothesis is our clean. We want to know if the carb content in the Candies differ. So that stuff a Now we're gonna go to step be, which is finding are critical value, which we use the H table to dio. So the sample size for chocolate is 13 and the sample size for a non chocolate is 11 But we don't know which we're going to use for a numerator and which we're going to use for a denominator until we figure out the variances and because they gave us raw data, we can use an Excel spreadsheet to go ahead and calculate all that for us. So here's the Excel spread she. So the first thing we're gonna do is we're gonna calculate the standard deviation for each category and then to get the variance. All we do is we take that number and we square. So the variance for our chocolate is 42.23. And for non chocolate, it's 1 25.45 So are non chocolate, because this very intense larger. That's what we're gonna use for our numerator and because of degrees of freedom is our sample size minus one. It's gonna be 11 minus 2010. T minus one equals well So this is the degrees of freedom for our numerator. And this is the degrees of freedom for our denominator. And as we saw before, the variants for our chocolate was 42.23 on our variants for our non chocolate was 1 25.45 So, yes, this one is both in our numerator. So let's pull up the age table for our 0.5 off the level and again, we're gonna be using 10 degrees of freedom for numerator and 12 degrees of freedom for our denominator. Just a slider cursor till we get to tense are critical values to 0.75. All right, so that's step be and Step C is computing our test value. And to do that, we're going to be using our variances from our samples. So the variance of our chocolate waas 42.23 4 hour non chocolate it waas 1 25.45 So what we're gonna do is we're going to put our larger variance over top are smaller variance, and I'm just gonna go ahead and pull up that Excel spreadsheet again so we can go ahead and calculate it. So larger variants divided by the smaller variance. 2.97 So that is our answer or our value. So that's our test value. Step D now is making a decision based on our test statistic and our critical about so are how was 2.97 and are critical value. Waas 2.75 So what we want to ask ourselves is, Is our cow greater than are critical? And if so, that's when we can reject are no hypothesis. So is 2.97 greater than 2.75? Yes, it ISS. So we are going to reject Arnold hypothesis. So that stuff D and step ease our last stuff and it's summarizing our results. So because we rejected Arnold hypothesis, there is enough evidence to support that the variance and our content between chocolate and on chocolate can. All right, so in this video we looked at an example, performed an up test using need judicial, traditional method of hypothesis testing. And we did reject Arnold hypothesis. Um, I hope you guys enjoyed this video. Learned a lot and I can't wait to see you next time

We have a sample of six students who assembled a toy once and was timed. And then they assembled a toy a second time and were timed again. And we want to test if at a significance level of 0.1 if there is evidence to support the case that it took them less time to assemble it the second time, where has to do this test using the P Valley method, and then we're as to construct a 99% confidence interval for the difference in the means. So we're doing this test at a significance level of 0.1 So first step is to identify our hypotheses. So then no hypothesis is that the mean of the difference in times is zero or that there was no change. And the alternative hypothesis is that the mean of the differences is not zero. In fact, it should be creator than zero because we want to test whether the times were reduced on the second construction of the toy and from the problem we have an is equal to seven. So since we're using the P value method to test this, we can go straight to calculating our test value. That's given by this formula. And so we have and from the question. And we need to find the average difference in times and the sample standard deviation of the difference in times. So I've calculated these parameters in Excel. So I copied and pasted the sample data from the question here and in this column. I've calculated the differences in times or trial one minus travel to and here I've calculated the average of the sample differences. This is the average for column J, and here I've calculated the sample standard deviation for the differences. So now we have 11.86 divided by 10 point 37 over the square root of seven, and this comes out to 3.0 to 6. So now we can go find RPI value, knowing that the degrees of freedom is in minus one and the significance level is 0.1 So we have six degrees of freedom and our test statistic was 3.0 to 6, which would be between these two values, which means that RPI value, which is the area in one tail, is between 0.1 and 0.25 And we used the area in one tail because this is a right tailed test based on our alternative hypothesis. For now, we can compare RPI value to Alfa and we can see that the P value is greater than Alfa and therefore we would fail to reject the no hypothesis. And we could also add a statement of our conclusion, which is to say, at the significance level of 0.1 there is insufficient evidence to support the case that trout to was faster than travel one for assembling the toy by these Children. So that's the first part of the question, and the second part was to construct a 99% confidence interval for the difference in the means. So when 99% confidence interval is of this form, it's the mean difference, plus or minus cease of T's of Alfa over to times a sample standard deviation over the square root of the sample size. And so we already have almost all of these values calculated. The only thing we need to find is T's of Alfa over to, and that is based on significance level of 0.1 Well, it's a 99% confidence interval, so we have an area of 0.1 in two tails and the degrees of freedom is six. So if we go back to the tea table now, we're looking for an area of 0.1 in both tails and so T sub ALF over two is 3.707 So this comes out to 11.86 plus or minus three point 71 times 10.37 over the square root of seven. And this gives us a confidence interval from minus two point 67 to 26 0.38 And so we can write that as such and these units earn seconds and we can communicate concluding statement about our confidence interval. We can say we are 99% confident that the true difference in assembly times between the first and second trials is between negative 2.67 seconds and 26.38 seconds

Whether the every sick day is less than 10 Given the significant level up A is 0.5 soul estatal hypothesis. So it's not is new Eagle to 10 and bishop one is mu less than 10. This is our claim. So this is left, oh test To get the test value we need to buy the exper which is zero plus all the way to NY divided by number of sample which is Explorer is by poi 03 So the test value will be five boyo three minus 10/3 30.63 divided by squared off 40. So this give me the value of nearly a point a poise 66. So let's look at the tea table here we only have the small it value of Z isn 83.49 at the p value our poi 0002 So I only bring it here and then So a Z is native a 0.66 which is even smaller than 83 points 49 we have the P value would be less than the P Valley over here, which is 0.2 and this number is less than Alfa, which is 0.5 So RPI value is around these areas very, very small. So this mean we go into reject the no hypothesis ish non. So we're gonna come call that there is enough evidence to believe that the manager statement is right. That is the every sick day is less than 10. Given that a significant level Alfa is 0.5

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.


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