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A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from t...

Question

A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is $49.50$ minutes with a standard deviation of 3 minutes. Assume that the drying times for these paints have a normal distribution.a. Using the $1 %$ significance level, would you conclude that the company's claim is true?b. What is the Type I er

A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is $49.50$ minutes with a standard deviation of 3 minutes. Assume that the drying times for these paints have a normal distribution. a. Using the $1 %$ significance level, would you conclude that the company's claim is true? b. What is the Type I error in this exercise? Explain in words. What is the probability of making such an error?



Answers

The mean television viewing time for Americans is 15 hours per week (Money, November 2003 ). Suppose a sample of 60 Americans is taken to further investigate viewing habits. Assume the population standard deviation for weekly viewing time is $\sigma=4$ hours.
a. What is the probability that the sample mean will be within 1 hour of the population mean?
b. What is the probability that the sample mean will be within 45 minutes of the population mean?

We're going. The following data set in this problem and what we're trying to see is if the mean drying time is the same for each type of paint. So let me just write down our no little alternative hypotheses here or no hypothesis, this plan will work. Yes or no hypothesis. Is that the mean for each treatment group is the same. So we have four treatment groups that will have four means, and they're all the same. That is our no hypothesis And our alternative hypothesis is that not Oh, the means are the same means for each treatment group are the same. So in order to figure this out, we have to take in a nova test. And to do that, we use what we call in a nova table. This is a neat way to capture all our information. The first thing we're going to do is find the sum of square of our treatments. But in order to do that, we have to calculate, um, the sample mean for each, um, each population here. So the sample mean for paint one is equal to the sum of all of these divided by the number of elements. So we get a sample mean of 13.3 Simple mean for two is equal to 1 39 Oh, sorry, I meant one 33 13.1 33 1 39 If these cars will work, Sample mean for the third population is 1 36 sample mean for the fourth population is 1 44 So using this information we're going to find or grand mean so our grand mean is equal to the sum of each individual mean divided by the number of individual means. We're just taking the average of the each sample mean and we get a value of 1 38 So now, using this 1 38 we can come up with a sum of squares for a treatment group which is equal to the sum from J equals one. And J is just a counter to the number of treatment groups that we have times number of elements in each treatment group times the difference between each sample mean and our grand me squared. So this is going to be equal to five times 1 33 minus 1 38 1 33 minus 1 38 squared plus five times 1 39 minus 1 38 squared plus five times 1 36 minus 1 38 squared plus five times 1 44 minus 1 38 squared. And we get a sum of squares of our treatment of 330. So in here we can update this value. This is 330. The next thing we're going to do is find the sum of square of our treat of our total. And this is equal to the difference between each individual value and the grand mean squared. So we're going to take the difference between 1 28 and our grand mean of one 38. So 1 38 or 1 28 year minus 1 38 squared 1 37 minus 1 38 squared 1 35 minus 1 38 square. And we're gonna do that all the way until 1 53 minus 1 38 squared. And now we're gonna take the sum of all those differences. So we get a sum of squares or treatment of our total. Sorry to be, uh, where is it? Some squares of our total to be 10 22. So beautiful, you 10 22. And now we're going to find the sum of square of our error, which is equal to the sum of squares over a total minus the sum of square of our treatment that is 10 22 minus 6 92 It's and then that is equal to Bert. Sorry, 10 22 minus 3 30 which is equal to 6 92 So we have a sum of squares of errors of 6 92 and now we're going to find the means square where treatment mean square of our air. And in doing so, we will find the degrees of freedom for each. So the mean square of our treatment group is equal to the sum of squares of our treatment group over the two years of freedom for Treatment group. And as we know, the degrees of freedom over treatment group is equal to the number of treatment groups minus one. So we get 3 30 and because we have four treatment groups with 3 30 over three, which is equal to 1 10 So we have a degrees of freedom of three mean square of 1 10 and the mean Um square for error is equal to sum of square of our squares of our error over the degrees of freedom for our error, the degrees of freedom for our error is equal to the end of the total minus the number of treatment groups which is equal to We have a total number of elements of 20 minus four treatment groups. I get this 20 by counting the number of each individual items of 12345 I have four rows, so that's 20 elements. So 21 is for is equal to 16. So I have a sum of square errors. To be 6 92 over 16 is equal to 43.25 You know him? I know the table to this here is equal to 16 and this is equal to 43.25 and I have to find the degrees of freedom for our total degrees of freedom. For a total I simply the end of our total. The number of elements we have minus one. And as we just discovered, our we have 20 total elements and that minus one is equal to 19. So we have a degrees of freedom for a total between 19 and now we're going to find a F statistic thief. Statistic is simply the means square of the treatment over the main square of our air is equal to 1 10 over 43.25 which is equal to approximately 2.54 And now, using Excel, we confined a p value for that O. R P value is going to be equal to and this is a formula. If d i s t our f statistic, our the degrees of freedom for our treatment. Just three years of freedom for our error, which is 16. And then this yield the value of approximately 0.93 And now we're asked to test our 0.93 p value against our Alfa of 0.5 So because 0.93 is greater than 0.5 there is not sufficient evidence to reject the no. So what that means is that we do not have sufficient evidence to support the claim that the means for the four treatments are not equal

So this question is asking us to figure out if Marcella takes a shower for seven minutes. Is that unusual? So, basically, is that an out liar? So how do we figure out out flyers for that? We first need to know what distribution The data's following. And they've told us that her shower data follows a normal distribution. Um, so basically a normal distribution has a school table called See Table for if you know the Z score. Um, any area to the left off it is the probability of that happening. So let's first find out the Z score off her, taking a seven minute shower So the formula for a Z score is equal to X minus mu over the standard deviation. Andi, In this case, the X that we're trying to figure out is seven minutes. Her usual shower time is 4.5 minutes. Onda the standard deviation of 0.9. If you plug all this into a calculator, you kid 2.78 approximately, Um, now we need to find out if that's more than her usual. So we need to find the Z scores for, like, what's her usual range on to do that, we will first find out disease score off the 25th percentile. So the probability that X is less than 0.25 So this whole curve, the area under this whole curve is equal to one. And, um, if the area 0.25 under the curve, uh are obviously that corresponds to a 25%. Probably so to find out the Z score of that, we need to look at this whole table and find where 0.25 is. Um, it's kind of like a lot of numbers, but they all follow pattern. And soon you'll find that in this 0.67 row, the numbers are closer to 2.5. So 0.25 on and this is where is your 0.25 would line between, um this number is closer. So let's go with this. And I would figure out which call him this lies and you just go up this way. Find out it's in 0.7 So the problems are Z score. Ah, for this probability is equal to Z is equal to negative 0.6 plus 0.7 is equal to negative. So porn 67 One thing to notice here is that this is negative 0.6 But this is not just a regular edition. This is just, like, more off appreciate position, a precision maker. So you just really add what this value is off to that decimal place. So it's not actually negative. 0.6 plus 0.7 It's just negative 0.6, and then you're giving it more precision by adding that 0.7 part. So that's a Z score. The precision matters because these probabilities, um, small numbers make a big difference. So that's a Z score for the 25th percentile, uh, disease score for the 75th percentile. So it let me p off X greater than 0.75 We don't have to look at the table, and this table wouldn't even tell us that, because it's for values up to 4.641 so we wouldn't even get 75 from here. But that's because this table is not telling us the area to the right of the curve. It's only telling us the area to the left and it's only telling so certain part, so we can just use some cleverness. Figure out that so this is a symmetrical curve. Um, if this is 25% of the area, um, this should be 25% of the area to the right, which is the area greater than 75. So and since this is centered on zero and this is the metric curve, the negative value of this should be the positive value we're here. Which would be in our case, we got negative 0.67 So there should be positive 0.67 So we found, um, the ranges for the Z score. So anything between negative 0.67 to 0.67 is the Z score off her regular shower times. Now we need to figure out if, um, her 2.78 see score is an outlier. So we take these 25th percentile and 75th percentile, and we calculate the inter quartile range first. So that's like you are. The formula is Q three minus. Q one is equal to well in this case is just 0.67 plus 0.67 and that is 1.34 Um, to find out the out liar, we use a formula that and uses like you are, um, so anything that is more than 75 more than 75 percentile, plus 1.5 times the I Q R, which in our case is 0.67 plus 1.5 times 1.34 and this is equal to 2.68 So we found this number. Anything above this number is an outlier, uh, and then, um, the score for her when she takes a shower for seven minutes. This is for seven minutes. It's 2.68 So So it's 2.70. So 2.68 is lesson that therefore it's not that for taking shower for seven minutes for her is not liars. So let me quickly right That statement down since 2.78 is greater than 2.6 A. A seven minute faour is an outlier part B. So Part B is asking us, Look at the question real quick. Suppose we choose 10 days at random and record length of Marcella Showers each day. What's the probability that our shower time of seven minutes or higher on these two of the seven days sure work s so we're gonna have to use some probability for this. But the first step is to find out the sea score. So we already did that in the first part. So we found out the Z score for for regular shower for seven minutes is seven point four or 57 minus 4.5, divided by 0.9, which is equal to approximately 2.78 Um and we just have to find out what is the corresponding probability off the shower happening. So we already noted, outlined from the first part. So basically, we have to find out what is, Ah, probability off X being greater than this amount being greater than seven, which is equal to the probability off Z score being greater than 2.78 So basically, this part is over here, and this part is over here, they're the same thing. But this is the language of X, and this is the language of C. They're the same thing. Um, since we are, table is only with negative values. Weaken use symmetry thing that we use in the first part and find out Z score being less than negative. 2.70. Ah, Now we have to go up and find the values. So 2.678 sorry to 0.78 would be between 2.7 and 2.8 on bond. We can use 2.7 and go to the 0.8 part on. So this is that on the eight part here to go down this row and hit this goal. This is our number 0.27 because we're trying to find the probability no z score less than that. There a 0.27 Um, that's a probability of ex being greater than seven. Andi Probability Off X being less than or equal to seven. Yes, the probability off a Z score less than 2.70. Sorry. Yeah, less than 2.789 I removed the negative sign. So it's different. And that would just be one minus 0.27 Which is it? Cool to 0.9 973 Okay, so we found the probability off something being greater than seven and the probability of something being less than seven. Um, obviously the probably of something being less than seven is much higher because she very rarely takes a shower that's longer for seven minutes longer than seven minutes. And now we're asked to find the probability in this question that at least two days, at least two days out of 10 I was gonna write this slash. But it's not divided out of 10 are greater than seven. So this probability at least two days off out of 10 is greater than seven. So there's two ways this could happen. Um, week. Sorry, there's there's like so many ways this could happen either. The first day could be seven minutes long and heard they can be set. Seven slowing or the first they can be. Assignments on the fourth day can be seven, as long basically goes on and on and on their so many combinations, but based the quicker ways to take all the possible combinations of her taking a shower. Um, lesson earlier than seven minutes long on finding out the probability that none of her days zero off 10 days is greater than seven. Andi also subtracting from that The probability off one of 10 days is great. It's having good part of me. So that kind of encapsulates this exact same probability. So at least two days out of 10 being greater than seven is the same as all the days except the days where, UM, none of the days were greater than seven on. Only one of the days was greater than seven. So we just have to use these two probabilities. Andi, subtract that from the hole and we can find the same question we were looking for. That's what this question we're looking for. So that would be one minus zero off, 10 days great and seven. So we found the probability off that this is the probability on one day her shower is great and seven, and we multiply that with the number of days we have. So that's the same as taking the power off it. So 0.9973 raised to 10 because for each day we have to multiply. Um, that probability minus 10 times zero point nine 973 race to nine times the probability that one day was greater than seven. That was very small probability. So that's number we're gonna use the reported 00 to 7 on that is going to be equal to zero point 000323 So deconstructing this might be a little difficult because it takes in several probability rules. But basically, um, the rule that were user the p of A and be is equal to p of a times P f B on the p off, not a is equal to one minus p of a. So we used the one minus p of a part here when we did one minus p. Of all the stuff on the p of a types pf be is when we did the multiplication here. So the final part of this question sorry for being so long find the probability that the mean lent off her shower time on these 10 days exceeds five minutes. So this is not gonna be that, but this is not gonna require too many laws. We just first find out the Z score off her shower, taking five minutes, and that is five minus 4.5 over 0.9 divided by Route 10 just applying the formula Disease core, which is 1.76 And now we use the P off X greater than five. So that was a Z score for fire, so we already know that. So this was a Z score for five. So we already know how to convert this to Z language. So p off the sea, being greater than 1.76 And since our table only has negative numbers, weakened issues, the fact that it's symmetrical 1.76 So we need to find disease probability of Z score being less than 1.76 Go back to our table. Uh, being less than 1.76 So it's 1.7 is over here on. We find there are six here we go down this column and we get this number. Sarah 0.392 That's a probability

Let's start this problem by writing down what we know. So we have the average wait time in a doctor's office, varying the standard deviation which we could define as Sigma is 34 minutes. We're going to randomly select 30 patients, so our sample sizes 30 and the standard deviation of those wait times is going to be 4.1. We have a belief or a claim that the doctor is making, that believes that the variance, which should be Sigma squared, is greater than originally thought. So we're going to be running a chi Square test of a single variance on this. And in order to do so, we're going to have to write a hypothesis and a no hypothesis now because our belief or our claim has to do with our variants. We're going to have to take our standard deviation statement and turn it into a variance statement. So are variants statement since variances the square of the standard deviation. We could say that Sigma squared equals 3.4 squared. So now that we have that variant statement, weaken, fulfill the rest of the claim and say that Sigma squared is greater than 3.4 squared. So when we're writing our null hypothesis, it's always a statement of equality. So we're going to say that Sigma Squared is equal to 3.4 squared, and our alternative is based on our claim, which is going to be that Sigma squared is greater than 3.4 squared now, Ultimately, we're trying to find our p value. And in order to find RPI value, we will have to first find our test statistic. So our chi square test statistic is found by utilizing the formula n minus one times the standard deviation of the sample divided by the standard deviation of the population. So we have all of the information that we need. So let's substitute our values in. We know that Ennis 30. So we'll have 30 minus one. We know that s is 4.1. So we're gonna multiply that by 4.1 squared and we know that Sigma was 3.4. So we're going to divide that by 3.4 squared. And when we do that, the chi Square test statistic would be 42.1704 Now, when we run a chi square test of a single variance. It could be a left tail test, a right tail test or a to tail test. And that determination is made based on our alternative hypothesis. And since we're talking about the variants being greater than the 3.4 squared, therefore it's going to be a right tailed test. So to find our P value, if we were to draw our Chi square distribution and we now know it's a right tailed test, R P value is going to be found right there. So our P value is going to be the probability that our chi square value is greater than the test statistic we just calculated now when we drew that chi score distribution, the shape of the distribution is dependent on the degrees of freedom and degrees of freedom can be found by doing n minus one. And we knew we randomly sampled 30 individuals, so our degrees of freedom is 29. That 29 is also going to be the average of the chi square distribution or the mean of the chi square distribution. So we know that that 29 value would be slightly to the right of the peak and the beginning of the graph is always zero. And what we're trying to do is replacing this test statistic on here as well. And we're trying to find the likelihood or the probability that were greater than that value. In order to do so, I recommend using your graphing calculator and accessing your chi square cumulative density function. And when you use that function, you have to provide the lower boundary of the shaded area, the upper boundary of the shaded area and the degrees of freedom of the curve. So in our situation, the lower boundary is that test statistic the upper boundary. Keeping in mind that that right tail extends infinitely, we're going to pick a very large number 10 to the 99th Power and our degrees of freedom we calculated just before to be 29. So I'm going to show you where you confined the cumulative density function on a Texas Instruments calculator. So I'm gonna bring our calculator in, and we're gonna hit the second button and the bears button, and it happens to be number eight on my menu, and we're going to put in the lower boundary, the upper boundary and the degrees of freedom, and we're getting a P value of approximately 0.542 So what is that telling us that is telling us that this area right here is 0.542 or accounts for approximately 5% a little over 5% of that chi square distribution. So just to summarize, we were trying to determine the P value for this scenario, and it ends up being 0.542

Let's start this problem by writing down what we're given. Suppose an airline claims that flights air consistently on time, with an average delay off at most 15 minutes, so that means mu is less than or equal to 15. Then there is a claim being made that the average delays so consistent that the variance, which would be Sigma squared, is no more than that would be less than or equal to 150 minutes, doubting the consistency part of the claim. So we're only doubting we're not doubting the average were doubting the, um, variants. A disgruntled traveler calculates the delays for his next 25 flights or sample size is going to be 25. The average delay, so that would be an X bar of those was 22 minutes with a standard deviation. So that would be a sample standard deviation of 15 minutes, and we need to complete the statement. A sample standard deviation of 15 minutes is the same as a sample variance of how many minutes. So since variants is classified as standard deviation squared, we're going to take the sample standard deviation, and we're going to square it and get a result of 225 so we could fill in that blank with 225 minutes.


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