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The data given to the right includes data from 43 candies and oi them are red. The company that makes the candy claims that 29%0 of its candies are red Use the samp...

Question

The data given to the right includes data from 43 candies and oi them are red. The company that makes the candy claims that 29%0 of its candies are red Use the sampl data 10 construct 9096 confidence interva estimate oi the percentag oi red candies_ What do you conclude about the claim oi 2990?Weights Blue 393 926 823 886 946 0.842Sample Bag of CandyYellom 836 836895 805893Construct 090 confidence interval estimate of the population percentage of candies that are red.721 7940.784795 782(Type an

The data given to the right includes data from 43 candies and oi them are red. The company that makes the candy claims that 29%0 of its candies are red Use the sampl data 10 construct 9096 confidence interva estimate oi the percentag oi red candies_ What do you conclude about the claim oi 2990? Weights Blue 393 926 823 886 946 0.842 Sample Bag of Candy Yellom 836 836 895 805 893 Construct 090 confidence interval estimate of the population percentage of candies that are red. 721 794 0.784 795 782 (Type an integer or decimal rounded to one decimal place as needed) 959 878



Answers

Use the indicated data set in Appendix $B$. Data Set 27 "M\&M Weights" in Appendix B includes data from 100 M\&M plain candies, and 19 of them are green. The Mars candy company claims that $16 \%$ of its M\&M plain candies are green. Use the sample data to construct a $95 \%$ confidence interval estimate of the percentage of green M\&Ms. What do you conclude about the claim of $16 \% ?$

So in this problem, we, um we have 16 backs of Candies and, um, the weight of each bag is recorded. So we have 16 samples and total. Yeah. So the number off some holes is on equal 16. And, um, the sample meanest too. So we have X bar. Because who? Andi. The population standard deviation equals 0.1. So said my once 0.1. I don't Andi, um, the sample standard deviation equals 0.12. Yeah, And here we have the answers for part A. Yeah. Now, part B asks us to define the random variable X um, in this problem, we're concerned about the weight of each back. So, um, access defined to be the weight of the back. Andi in part c, we wanted to fight. Um exp are so X bar is the the mean of X, which means, um the average Sorry, the average weight off the backs and part D. Um, yeah, we want to know that the distribution were concerned about for this problem. Onda. We already know that the bags the weight of the bags is subject Teoh a normal distribution so we can write it as and um packs far sit my divided by some hole number on. We already know the values. So if we substitute all the values into it, we have to and 0.3 now in part E where, um, we're asked to compute the 90% confidence interval for the, um, population means so the, um in order to In order to calculate the confidence interval first we need to compute the Arab bound and they are bound equals two. The score, um, times siglo divided by square root a pen. And we need to know Alfa Alfa equals Guan minus 90% and equals 0.1. Andi Alfa, divided two by two equals Cuando five. Yeah. So, um now way. No, they are bound and they are about equals. You can compute a Cisco or using a calculator or, um, computer program, like are are met lab. Um, the Z score equals 1.645 times 0.1 square root of 16. On this issue calls 0.41 and back. Um, the confidence interval is exper class miners are about in this case, the confidence interval is 1.9 five nine and 2.41 So this is the This is the 90% confidence interval we finally get. Um, And if you look at this graph, this is a standard normal distribution on Dhere is it's mean. And, um here is the Z score and, um, the area, um, at the right of the Z score, um, equals Alfa divided by two. So that that is the definition definition of the Z score. Um, so you can you can get this value by using a calculator or a computer program. So now we can look at part F. Um, let me raise hall of fees. So part of it is quite similar to who? Sorry. Okay, Yeah. Yeah. Part of it is quite similar to part eat. Except the confidence interval is, um, at a different level. Um, so we're asked to calculate, um, 90% confidence involved. So, um, first Alfa equals one minus 98%. This is 0.2 Okay, Andi. So Alfa, divided by two equals 0.1 And using the same formula and party, we can compute their abound. Which is the score at Plano, Juan. That's times Sikma divided by score. Word of on on these equals quiet 058 And, um, so the so Now we have the Arab bound and we can write the final confidence involved. Um, confidence interval is still x bar the sample. Mean plus minus Arab out. Yeah, and now it iss one point night for two. Yeah. Profound is 2.58 Yeah. So now we have a new confidence interval. Yes. And I'll, um I'll circle it. Okay, So if you compare these two, you can It is quite obvious that, um, this one is larger than this one. Um, because this this lower bound is smaller than previous one, and this hour bound is larger than the previous one. So, um, the 90% confidence interval is contained within the 98% um, confidence in her vault. So, um, so why is this one in part off Larger than that one in part E. This is because, um, the definition of confidence interval is that, um so for for for the interval in part of, um, this interval, it has 98% chance to cover the true purple prop population. Mean so. And for this one in party, it only has 90% chance to cover the true population means so of course, um, the one part of should be larger because it has a higher chance. Thio contain that the true me. Sorry. And and this is Thea answer for, um Targhee. Andi part age asks us to, um, explain the the confidence interval in part off. And are I already did. Um, I can repeat it again. Um, this interval is 3 98% confidence interval. And it means this interval has 98% chance tow cover the true population means and so that's all for this problem.

For this question, We've been given our sample sizes 14. Those are degrees of freedom is 13 since we're completing in 95% confidence intervals are Alfa level is 0.5 Now, since we're given the sample, we can calculate the sample standard deviation using this formula here on if you calculated, you'll see that the value comes to 23.83 Now we find critical that these guys were Eltham. I do and guys for one minute self over to I scored Alphabet do. In our case, the Sky Square 0.5 were degrees of freedom 13 checking the G table. You'll see that this where new comes to 24.736 guys were won by myself or by doing our case The Sky Square 0.9 some life with degrees of freedom 13 checking the G table. This value comes to 5.9 Now that we found the critical values, we can move on to calculate in the confidence interval for a population standard aviation, you know. So we use this formula here. This is the expression for the populations for the confidence interval for population standard deviation, something in the values we have. We get this expression here and simplifying this further, you'll see that the lower limit off the confidence and durable for a population standard deviation comes to 17.3 and the upper limit comes to 38.4. So this here is how you can lead the confidence interval for the standard deviation.

So in this problem, we want to test to see if the reported percentages by the company are actually true. Based on our data, so are no. Hypothesis is going to be that, um each of our proportions are equal to those reported by the company. So the proper proportion of population that's brown would be 0.3 portion of the population that's yellow would equal 0.2 proportion of the Eminem's population that is equal, that is red is equal to 0.2 proportion, That is, orange is equal to 0.1 proportion, that is green is equal to 0.1 in the proportion, that is, and blue is equal to 0.1. And now our alternative, I thought this is would be that, um, the percentages percentages reported by the company are not all the same. So we're given the following information in the problem. I just wrote this into this table here and now we have to find out the rest of our data. So the first thing I'm going to find is the expected frequency. The expected frequency is the probability of proportion given by the company times our end and our end is going to be equal to the sum of our observed frequencies and were also given in the problem that are in his 506. So in our expected frequency for the number of Eminem's that are brown would be 0.3 times 506 which is equal to um 1 51 0.8 1 51.8 and then it would be one A 1.2 101.2 50.6 and the restaurant just 50.6 because they all have the same proportions. And now the difference between the two would be or the difference would be the difference between the observed frequency and the expected frequency. So that would be 1 77 minus 1 51.8 So we get the following differences negative 12.6. And now we have Thio Square. This data and the reason we're doing this is because we are looking to find a chi squared goodness of fit and are to conduct a hypothesis test and the formula for our chi square goodness if it is the sum of, um, each of the square differences between between the, um given the actual frequency minus the expected frequency squared. Sir, I just want to get the indices the same. So it would be the expected frequency minus for the actual frequency. Modesty expected Frequency square over the expected frequency. And we're going to take some of that for all of our categorical groups, so we need to find the differences first. So, um, our first difference squared would be 6 35.4 Yeah, for 92.84 92.16 to 13.16 and 1 58.16 And now we're just going to find the value of what we're going to some for Kai squared. And that would be on. I'm going to write the truncated, um, values approximately 4.18 approximately 11 point to nine. Approximately 4.87 Approximately 1.82 Approximately 4.21 in approximately three 0.14 And now, in order to find our chi squared, we're going to take the sum of all of these values, and we get it. Kai squared of 29.5 one. And now with that kai squared of 29.51 We need to figure out our degrees of freedom so that we can come up with our P value. So our degrees of freedom, simply the number of categories that we have minus one. And in this situation we have six categories, so we have six categories brown, yellow, red, orange, green and blue. So we have five years of freedom. So our P value given that chi square is equal to 29.51 we have five degrees of freedom is equal to, um, less than point less than 0.5 So compared to an Alfa of 0.5 and the P value less than 0.5 our P value is less than 0.5 which is less than 0.5 are Alfa. So therefore, we reject the no hypothesis and what does that mean? That means that there is sufficient evidence to reject the claim that the companies percentages are correct

This question asks us about the distribution of colors in bags of Skittles. Now the claim is that we're trying to disprove. The claim is that every color is equally likely. So are no hypothesis is that it's true that the claim the probability of red is equal to the probability of orange, which is equal to the probability of yellow, which, of course, is equal to the probability of green and is equal to the probability of purple. And since he's the only five colors, they all some upto one each probability is just one of her five or 0.2. Now, where we have some data that we're hoping to use to disprove this no hypothesis, Um, and that is a small distribution of about 106 Skittles. So we're gonna use a chi squared test here to test the likelihood of this null hypothesis. Um, there test the likelihood of our data given the not about This is so. What we have is a list of observed values for Redd with 18 red, 21 oranges, 23 yellows, 17 greens and 27 purples. So those are observed values. Now, what are are expected values. Well, we have a total of 101 106 Skittles, and each color has an expected frequency of 1/5. So each expected value value will be 106 divided by five. So 21.2. 21.2 all the way down will be are expected values. Next. We want to find the difference between are expected values and are observed. So oh, minus e observed minus expected is gonna be first negative. 3.2, then negative 0.2. Um, then we have 1.8 negative. 4.2 and 5.8. So from here, we next want to square this, um, because that is part of our chi square, statistically to square them and then divide by the expected value. So when we square each one, we get 10.24 0.43 point 24 17.64 and 33.64 And finally, um, we get the piece that we want to add up. Observed minus expected square over expected will divide each one by 21.2 to get this'll 0.483 0.2 I'm around in a little bit 0.153 0.8 32 and 1.587 Now I'm rounding these values a little bit, but I'm not rounding the values that I added up all together. When I added all of them up the full values I got Ah total of 3.566 And now that some of the observant is expected squared, divided by expected, that is our test statistics that's gonna be chi squared star as her test statistic towards noting here that we have four degrees of freedom, right? We have five categories so new our degrees of freedom. It's gonna be five minus one or four. So first we want to do the P value approach. The P value approaches refined the probability of a chi squared random variable with this many degrees of freedom being greater than our test statistic. So RPI value is going to be the probability that a chi squared random variable with four degrees of freedom is greater than chi squared star or 3.566 And when we played this into a calculator or look it up in a table. What we're gonna find is that RPI value here is actually much greater then 0.5 which is our Alfa level. The P value is 0.548 Of course, this is greater than 0.5 are suggested Alfa level. So we fail to reject are no hypothesis. Our next bit we want to use the classical approach. The classical approach involves finding a, uh, critical value. Now, the critical value for this is gonna be chi squared with four degrees of freedom at a 0.5 level. And when we use table eight in our appendix, we find the 0.5 on the right and four degrees of freedom. And when we find the intersection of that row in that column, we get that the critical value here is 9.49 Now, since our test statistic I scored star is much less than 9.49 we know that we have again failed to reject the knowledge prosthesis. So what does it mean that we failed to predict In all hypothesis, it means that we don't have enough evidence to conclude that there is an unequal distribution of colors, so we might as well continue believing that each color is distributed equally and that's your final answer.


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