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You may need to use the appropriate appendix table or technology question to answer this A random sample of 64 students at a university showed an average age of 23...

Question

You may need to use the appropriate appendix table or technology question to answer this A random sample of 64 students at a university showed an average age of 23 and a sample standard deviation of 5 years. What is the 95% years confidence intenral for the true average age of all students in the university? 018.0 to 28.0017.0to 29.0021.8 to 24.2022.0to 24.0Question 2 (1 point)LutenYou may need t0 use the appropriate appendix table or technology to answer this question: The manager of a grocery

You may need to use the appropriate appendix table or technology question to answer this A random sample of 64 students at a university showed an average age of 23 and a sample standard deviation of 5 years. What is the 95% years confidence intenral for the true average age of all students in the university? 018.0 to 28.0 017.0to 29.0 021.8 to 24.2 022.0to 24.0 Question 2 (1 point) Luten You may need t0 use the appropriate appendix table or technology to answer this question: The manager of a grocery store has taken random sample of 225 customers Tne average length of time it took these 225 customers to check out was minutes: Ibis known that the standard deviation of the population of checkout times is minute The standard error is 0.0667. Find the margin of error; with 0.95 probability. that the sample mean will provide 0.131 01.307 00.110 031.097



Answers

Use formulas (5) and (6) or (5) and (7) to solve Problems 22, 23, 24, and 25. To use formulas (5) and (6) to evaluate the sample mean and standard deviation, use the following column heads: $$\begin{array}{l|l|l|l|l} \text { Midpoint } x & \text { Frequency } f & x f & (x-\bar{x}) & (x-\bar{x})^{2} & (x-\bar{x})^{2} f \\ \hline \end{array}$$ For formulas (5) and (7), use these column heads: $$\begin{array}{l|l|l|l|l} \text { Midpoint } x & \text { Frequency } f & x f & x^{2} & x^{2} f \\ \hline \end{array}$$ What is the age distribution of adult shoplifters (21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution: $$\begin{array}{l|ccc} \hline \text { Age range (years) } & 21-30 & 31-40 & 41 \text { and over } \\ \hline \text { Number of shoplifters } & 260 & 348 & 287 \\ \hline \end{array}$$ Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use 45.5 as the class midpoint.

Now let us look at this question. This is what we did in the last question. Let us read this one again. We're using the desert one more editor in appendix B, which includes 1 40 cent randomly selected adult females. And those ages have a standard deviation of 17.7. And here they're saying we able to assume Sigma also equal to 17.7. So in this case, this value is going to be 17.7 17.7. Okay. How many female statistics students ages must be obtained in order to estimate the mean age of all female statistics students. Okay, we want 95% confidence and that the sample mean it's within 1.5 years off. The population mean it's me. This is correct and we want 95% confidence this time. So this value of Z Alfa by two is going to change. What is going to be Alfa by two? It is going to be 0.0 to five and alphabet to Z Alpha by two is going to be 1.96 This value turns out to be 1.96 in this case 1.96 1.96 I just have to substitute the values I have Sigma I have zf of I do and I want to find out the value off n So the value of end is going to change here. And after doing this calculation, the value off and turns out to be 4815 and is equal to and is equal to four eat one fight. My N turns out to be 4815 And the next question is, does it seem reasonable to assume the ages of female statistics? Students have less variation? Yeah, you can say that the sample size is pretty large here, and we Can anybody see that by looking at this analysis?

All right? So here we have analyzing Uh huh fast food restaurants wait at the drive through. We have the mean is 59.3 seconds. That's that's fast ripping through there. It's like I'm sorry. Some of these these things are funny. I'm just imagining the car going through there that fast and you don't even break. It's like you drive through, you shout at the window, there's your order through the speaker and then someone tosses you drive by and they toss you the food and tossed the money at them. Mhm. Yeah. Uh But you have to slow down sometimes. Sandy ph is 13.1 seconds. Sometimes if you take a moment to really think about some of these problems that we do in these statistics courses they I think you kind of party. Uh But apparently it's happened. Uh Someone did some research and found this. So there you go. Alright. So they found this distribution was actually skewed, right? So here's some here's it would be like this would be skewed to the right, like So that's where you have a few on this. Maybe maybe it's like this it's like this bulge here on the left. That as time increases this way times going up. That's confused, right? So you have those times when you're making the order. I don't know what happens. It just takes a long time. For whatever reason, maybe they have long long lines of people on this one day or something like that during they drop a burger or fries or whatever, who knows what this takes longer. So it's skewed to the right so to do to obtain probabilities regarding a sample mean using the norm model, which is what we want to do because the normal model is just so nice to work with What sample size must be required. Well, the end must be greater than or equal to 30. And that's due to the central limit theorem. CTL central limit theorem, which says that as n increases the sampling distribution will Be approximately normal. As end gets bigger and bigger than 30. We can approximate it to be normal, but even bigger as this can be even more normal. So there you go. So then be uh They were trying to figure out if we sample 40 cars. Actually no, some 40 cars were sampled. Excuse me. And the meantime at the window is 56.9 seconds, which is less than the population mean? Uh huh. What's the probability that This 56.8 seconds or less? Uh Was it was like we're going to save this for later, but significant. What's the probability obtaining a sample mean? of 56.8 assuming this population is what it is. So, I'm using the the spreadsheet to do my work. But just to give you an idea of what we're looking for, is we want the probability That X. Bar is less than 56.8, little thicker Again. eight. Yeah, it looks much better. Next bar is less 56.8. And so we need to compute that into a Z. Score. But because it's a sampling distribution, we need disease X. Bar. And uh we need the way we do that is the beginning of the scores real Z. Thanks for equals X. Bar minus all these experts everywhere over the standard error of the mean. So X. Bar minus mu all over the standard deviation divided by Route. And that's called the standard error of the mean. And so for a size 40, I calculated it here. Here's the mean that were given population mean. The population standard deviation standard error. So we're gonna use disease score. And we're gonna I'll show the substitution is that we're going to do here. 56 point eight minus 59 3. All over 13. 1/40. And that's going to give us this number -1.2. So negative 1.2 oh seven round to the thousands place. And then we do the Z. Look up. And in a normal distribution, what that's gonna look like. Is this or here's the mean. Thank you, expert. And Well, I'm gonna show the value. Actually, let me show it is 59 three And 56 is gonna be somewhere here. 56 0.8. And this number that we do it. We would look this up. Uh look up negative 1.2 in the Z. Table. I'm using the this normal dysfunction in my spreadsheet. And That number .11 four. It's right here. Mhm. It's this area right here. So the answer is .114. That's what the probability is. Of getting the sample With the mean was 56.8 seconds. So pretty small. But we'll learn later on that. There was some parameters or some criteria we use. But right now do we think this is this is a new system? Because we're looking at, the manager implemented a new system and whether or not that system was effective. Um It doesn't seem that far off, but you could say it's a little bit far off depending how significant you want to be, but and then there you go.

For today's question, we'll be using formulas five and six from this section of the textbook. And since we are using these two formulas as stated in the question, will be using this chart to sulfur our answer to find the mid points. What we do is we take the lower bound of a class and we add the upper bound of the class and we divide by two. So for the first class of this data are midpoint is 5.5 doing the same thing for the rest of the classes, we get 15.5 25.5 and 35.5, noting that the 35.5 comes directly from the question. And now the frequency is just the number of data in each class until we had 34 18, 17 and 11. Again coming straight from the question. Now, for this column, we just multiply the numbers from this column with this column getting us 187 to 79 4, 33.5 and 3 90.5. Now for this column we need to find the sample mean for the frequency uh deviation. So what we do is we take this some of should be a some we take the sum of all of the data in this column here, and then we divide that by the number of data, and the total number of data will be the sum of all of the different frequencies. So the sum of all of these numbers is 12 90 1290. And some of these numbers is 80 so we do 1290 divided by 80 and we get 16.125 So this is our sample mean? Now using this number, we can fill in this section of the chart and it may be easier if you go this way, filling out for the rest of the chart instead of this way. So we'll find our uh midpoint minus our sample mean for this one. And we get negative 10.625 Don't let the negative trip you up because we're going to be squaring it. So we get 1 12.89 and then we'll get 3838.26 And so please note that all we do is square this value for this column and then we multiply this value by the frequency for this column. Now filling it out for the rest of the data. We get negative 0.6 to 5.397 point 02 And then we get 9.375 87 89 1494.13 And you get three and yet 19.375 squared is 375.39 And multiplied by the frequency of 11, we get 4129.29 Yeah. Mhm. Yeah. So now solving for our standard deviation, because you mean something for a standard deviation, s views from up here, Formula six. So we take the square root of the some of this column here, because as you can see this correlates to this, which is 9468.7. And then we divide that by end minus one and then we found to be 80 so it's 79. And then when we divide those and take the square root, we get 10.9479 and then finding our sample variance. Mhm. We know that this is just our standard deviation squared, so it's 10.9479 squared, which gives us 119.8576 Thank you for listening, and that is going to conclude.

Right. This problem is about a pharmaceutical company making tranquilizers, and we are given some data, and the problem is broken up into five different parts, so it's gonna take a little bit to go through it. Uh, the first part, part A is asking us to determine some statistics based on the provided data. So the first thing I'm going to do is I'm gonna bring in my graphing calculator and I'm going to hit this stat button and I'm going to select edit, and I'm going to place all my numbers into the graphing calculators. So I've put all the effective periods into the graphing calculator, and from there I'm going to have the calculator helped me determine all the statistics that I need. So in part, a subsection one, you're asked to determine the mean and to find the mean we're going to total up all the pieces of data, and we're going to divide by how many pieces of data there are. And when I total up those values, I get 22.6. I divide by the fact that there's nine pieces of data and I get approximately 2.51 now. Another way I could have done it because my calculator has the data I could hit, Stat. I can scoot over to calculate I could select one variable statistics and tell the calculator where you have stored your data and we have stored it. Enlist one. And by doing so, you can see now that the average is listed right there. X bar is 2.51 or thereabouts for part B of part two of this, it's asking us to determine the standard deviation, and it's denoted by S a backs. And we can get that right from our graphing calculator, which is recorded right here. And as I round that to three decimal places, it's approximately 30.318 in part three of part. Hey, we need to determine the value of N and N represents the sample size and the sample size. There was nine pieces of data, so it's gonna be nine. And in part four off part A. It's asking you to determine end minus 11 lower than nine would be eight, and there's a reason we're doing the end minus one, and it's going to be found when we do Part five of this problem gonna get rid of the calculator for the time being and let's go to Part B in Part B. It's asking us to define the random variable X inwards. So what does the X stand for? And in this particular problem, the X represents the effective length of time in hours, mhm for a tranquilizer. So how long does it last before it wears off? And then, in part C, we need to define X bar? And in this particular problem, it's the mean, effective length of time again. That's in hours for a tranquilizer from a sample of nine patients because there were nine pieces of data. So that stands to reason that there were nine patients. Part D in This is asking you which distribution should be used to solve this problem. So for Part D, we're going to use the students T distribution to solve this. And the reason we're going to use that is because we do not know the population standard deviation. We know the standard deviation of our sample. We calculated that using the graphing calculator, but we don't know the population standard deviation. And when we don't do that or We don't know that we use our T distribution. Now it's on to Part E and in part E. It comes in three parts and it's asking you to determine the confidence interval. It's asking you to sketch a graph, and it's asking you to calculate the error bound and in order to solve these, we can't solve them in the order that they're asked. What we do need to do is solve part three. First, we've got to find the error bound and the error bound of a mean using A T distribution will be solved using the formula T sub Alfa over to multiplied by s over the square root of n now S and n were defined in part A to find that t value and it's called the critical T value. We're gonna have to think about the confidence interval that we're going to determine. So we are trying to construct a 95% confidence interval for this population. Mean so the best approach is to draw out a bell curve and think about the fact that 95% of the bell is in that center, so that leaves 5% unaccounted for. That's out in the two tails. So the Alfa in this problem is 0.5. So therefore the Alfa over two is going to be 20.5 divided by two or 20.25 So what we're saying is we have 2.5% of the bell in the right tail and 2.5% or to five as a proportion in the left tail. Now, to calculate the tea that we're going to use in this formula, we're going to utilize our graphing calculator and we're going to use the feature in the graphing calculator called University. And when we use Unversity then asks for the probability or the area followed by the degrees of freedom. So let me take you to our graphing calculator and to access the university, you're gonna hit the second button and the variables button and select number four. Now, the probability that we're going to put in is what we put in those tails, which was 0.25 and the degrees of freedom we find by doing that end minus one that we answered in part A. So are degrees of freedom is going to be nine, minus one or eight. So we're putting in verse T we're putting our area that's in the tail, and then we're using eight Syrian type in our eight, and we get in answer a negative 2.306 So that value is corresponding to the T score associated with the location. That's to the left of the center of the bell. So we've got negative to negative 2306 here, and the one on the right side is going to be positive. 2306 And when we're using the formula and again, the formula is right here, that formula E B m error bound of mean equals, we're only going to use the value, the absolute value of those answers. So we're just going to use the 2.306 the S from part A was the 0.318 and the end from Part A was a nine serving divide by a square root of nine. And we're going to get the answer for the error bound of the mean, which is the part three of a letter E to be 30.2387 Well, now we're gonna go back and we're going to calculate the confidence interval. So now we're gonna go back and solve part one of part E. Now the confidence interval. What is the confidence interval? If we think of it on a number line, we have our estimate of the average here. And what we're going to do is we're going to add the error. We're going to subtract the error and it's then going to give us a new interval. So in interval format, we're trying to find a confidence interval at the 95 percentile. So we're going to take the average and we're going to subtract the error and then we're going to take the average and we're gonna add the error. So the average in this problem was two point five. And let's just go back up and do a double check. It was 2.51 So we're going to subtract the error that we got, which was this 0.2387 and then we're going to take the average. We're gonna add the 0.2387 And when you do that, you're going to get to 2713 as a low bound of our confidence interval, and we're gonna have 2.7487 as our upper boundary of our interval. So just to recap for part e sub three, our error was 0.2387 And for part one, which was the confidence interval is ah, low bound of 2.2713 and an upper boundary of 2.7487 And then finally, for this problem or this part, we needed to sketch the graph. So now we're going to do part two of Letter E. We're going to sketch the graph. So for part two, you're going to draw a bell shaped curve. We're going to show the 95% in the center we're gonna put the average in the center, which was 2.51 We're gonna put our lower boundary on the left boundary 2.2713 and we're going to put the 2.7487 on the upper boundary. And we're now talking about this confidence interval going from 2.2713 up to 2.7487 with a 2.51 as the average one more part to this problem we need to do part E or sorry, Part F and part f of this problem is asking you What does it mean to be 95% confident in this problem? If we were to sample many groups of nine patients, mhm 95% of those samples when you calculated the confidence interval would capture the true population mean of tranquilizer effective time. And I like to always give a visual of this. What do we mean by that? The visual would be, we have an average and the average was 2.51 and we select nine patients and we construct a confidence interval. And sometimes the confidence interval captures the mean and other times we calculate a confidence interval and the mean is an inside that interval. And what it's saying is, 95% of the time the interval is going to be capturing. That mean there's gonna be 5% of the time that the interval doesn't capture it, and it's to the left or to the right of the mean. So hopefully that helped you with E six parts of this problem


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