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Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?because the probability of getting sample mean ...

Question

Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?because the probability of getting sample mean of 0.8544 g or greater when 467 candies are selectedexceptionally small_

Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label? because the probability of getting sample mean of 0.8544 g or greater when 467 candies are selected exceptionally small_



Answers

Carbohydrates in Candies The number of grams of carbohydrates contained in 1 -ounce servings of randomly selected chocolate and nonchocolate candy is listed here. Is there sufficient evidence to conclude that the difference in the means is statistically significant? Use $\alpha=0.10 .$ $$ \begin{array}{llllllll}{\text { Chocolate: }} & {29} & {25} & {17} & {36} & {41} & {25} & {32} & {29} \\ {} & {38} & {34} & {24} & {27} & {29} & {} \\ {\text { Nonchocolate: }} & {41} & {41} & {37} & {29} & {30} & {38} & {39} & {10} \\ {} & {29} & {55} & {29} & {} & {}\end{array} $$

For this question, we are told that the number of fragments found in chocolate follows a Poisson distribution With λ equals 14.4 fragments For 225 g of chocolate on average. So that means on a program basis, the mean number of fragments program of chocolate is .064. And for part they were asked for the mean number of grams of chocolate analyzed until a fragment is detected. So let's say X. Is the number of grams of chocolate analysed between fragment findings. Since the number of fragments in the chocolate is a person distribution. The amount of chocolate in terms of grams analyzed between fragments is an exponential random variable. So we're we often think of exponential distributions pertaining to time. So, you know, inter arrival times, time between events occurring, but that dimension does not have to be time. It can be any other dimensions. So it can be distance or it can be mass as as is the case for this question. So here is a process where masses of chocolate are being analyzed and after so much analysis, fragment is found and then more masses analyzed. After analysis of so many grands. Another fragment is found. So the mean number of grams of chocolate analyzed until fragment is detected simply one over lambda. This is 15.6- five. For part B. We want the probability that there are no fragments In 28.35 g. So we can state this as the probability that we must that our next fragment is found after more than 28.3 g of analysis. 28.35 g. That's one minus the CDF, evaluated at 28.35. And that comes out to approximately .16-9 for part C. Were asked that if we consume seven 28.35 Grand bars of chocolate, what is the probability of no fragments? This is basically asking what is the probability that we go more than seven times 28 35 g before hitting a fragment. Remember the rate at which we come across fragments. Lambda remains constant across this entire process. So moving on from one chocolate bar to another, doesn't change anything about the processes. Just like analyzing the next gram of chocolate. So this is one minus the CDF at 798.45 g. And it comes out to approximately three point 05 Times 10 to the -6. So there's a very small probability Of getting through seven chocolate bars without coming across an insect fragment.

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.

In question. 14. They're looking for the probability of not red so I can add up everything That's not red, but I'm kind of on the lazy side. So I'm gonna take the probability of red, which is 0.2. And I know that everything has sat up one. So I'm gonna take one minus 10.2. And that means everything else has to add up to point. Hey, no, my probability of not ready.

In question 14. We're looking at the probability of not rid, which is one minus 10.2, cause I'm using the compliment rule and read his point, too, of any 1.8.


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