Intro Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to find and interpret a confidence interval for a population standard deviation. Here's our problem statement: A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 16 subjects had a mean wake time of 98.9 minutes and a standard deviation of 44.1 minutes. Assume that the 16 sample values appear to be from a normally distributed population, and construct a 90% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective? Part 1 OK, this first part asks us to find the confidence estimate as described in the problem statement. To do that, I'm going to bring up StatCrunch. And using StatCrunch we can actually get this but we have to put in a small caveat, and that is StatCrunch cannot directly calculate the confidence interval estimate for standard deviation. But it can do so for the variance. Because the standard deviation that we want is the square root of the variance, we have to make some adjustments to the numbers we put into StatCrunch and take out of StatCrunch. And I’m going to show you how this is done, so we’re going to get to this through the variance. To start with, I’m going to come up to Stat –> Variance Stats (because were going through the variance), we've only got one sample, and then I'm to select With Summary because we don't have actual data, just summary statistics. Here in my office window, it asks me for the sample variance. Notice we’re not given the variance directly in the problem statement, but we are given the standard deviation, 44.1. So I can get the variance from the standard deviation; the standard deviation is just the square root of the variance. So the variance is the square of the standard deviation. So in my calculator I’m going to put 44.1 (that's the standard deviation), I’m going to square it, and now this value here is the variance. If I copy and paste that here using Ctrl-V on my keyboard, I’ll copy that value right in there. That’s the variance. The sample size is 16. We want a confidence interval, and we want to construct the confidence interval estimate with a 90% confidence level, so I need to change this default value from 95 to 90%. I hit Compute! and here we have the limits of the lower and upper limits for a confidence interval on the variance. We want one for the standard deviation, so to do this we’re just going to take the square root of those limits, since the square root of the variance is the standard deviation. I’m going to bring my calculator back up, I’m going to take these values here, I’m going to put them back into my calculator, take the square root, round to two decimal places. So then my lower limit will be 34.16. And I do the same thing with the upper limit, and I get 63.39. Nice work! Part 2 And now the second part of this problem asks, “Does the result indicate whether the treatment is effective?” Well, the result we have is a confidence interval estimate on the standard deviation. Because standard deviation is the square root of the variance, it's measuring the variation that we have in the data values for sample set. So looking at the variation isn’t going to tell us very much about whether or not the treatment is effective. What tells us whether or not the treatment is effective is we look at, say, like the mean value and compare it with some standard or some threshold value, some requirement that we need to meet.
We don't have that threshold value given in the problem statement, a standard that we can use for comparison. And so there’s nothing that we can say from the actual confidence interval on the standard deviation. They’re just looking at variation with that. We don’t have any idea whether the treatment is effective looking solely at the confidence interval estimate on the standard deviation. So I’m going to look through my answer options. There’s only one that says it's not effective. I check my answer. Good job! And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video.
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Intro Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to find and interpret a confidence interval for a population mean with sigma known. Here’s our problem statement: Salaries of 42 college graduates who took a statistics course in college had a mean value of $64,300. Assuming a standard deviation of $16,562, construct a 99% confidence interval for estimating the population mean mu. Solution OK, because we know what sigma is (just the standard deviation for the population), we’re going to use a Normal distribution; we’re going to calculate z-scores. So to do that, I’m going to pull up StatCrunch. I could use the z-score tables, but my preference is to use StatCrunch. So inside StatCrunch, I’m going to go to Stat –> Calculators — excuse me, I’m going to go to Z Stats –> One Sample (because I’ve only got one sample I’m looking at) –> With Summary (because we don't have actual data, just summary statistics).
Here I’m going to put in the values that they give me here in the problem statement. There’s my mean value, there’s the standard deviation, there's the sample size. And now, because I want a confidence interval, I’m going to click the radio button here next to “Confidence Interval.” And we want a 99% confidence interval, so I need to change this default from 95 to 99. I hit Compute! É viola! There's my lower and upper limit for my confidence interval. So I simply slip those values here into my answer fields. I want to round to the nearest integer. Excellent! And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to find the sample size needed to estimate a population proportion. Here's our problem statement: The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 80% confident that his estimate is within six percentage points of the true population percentage? Complete Parts A through C below. Part A OK, Part A says we should assume that nothing is known about the percentage of adults who heard the brand. So the first step we’re going to have to take to calculate the sample size that we need is to find the critical value. To do that, I’m going to open up StatCrunch so I can access the calculator inside StatCrunch. I could also do this with the z-score tables, but I'm just going to use StatCrunch since it's my preference. I’m going to pull up the Normal calculator. And we want the standard Normal distribution; that's the default here in the Normal calculator. So then I want the two-tailed critical value, so I’m going to click the Between option. And then here in the percentage [field], I’m just going to put in I want 80% confidence. So that’s 80%. I hit Compute!. There are my two critical values; I really only need the positive one (1.28), so that's what I'm going to use. Now that I’ve found the critical value, I can actually substitute into my equation. Here's my equation for sample size, and I just substitute in what I have. So 1.28 was the critical value we just found. This is where we know it's a two-tailed area — z-alpha-over-2. Alpha over two says we want to split alpha amongst the two tails of our distribution. So that’s how I know it's two-tailed. So that's my critical value 1.28. We don't know anything about the percentage of adults who part of the brand. In that case, the most conservative percentage that we can collect for p-hat is going to be one half, which then means q-hat is also one half. ½ times ½ gives us 0.25. And then we want to be within six percentage points. So notice how we write that here — 6%, six percentage points of the true population percentage. Now that I've got my numbers substituted into my equation, all I need to do now is just calculate that out. We get 113.7 and the seven is repeated off into infinity. We’re asked to round up to the nearest integer so we get this partial person counted for. Since we can’t really count partial people — we’re only counting whole people — we’re going to round up to the nearest integer. That gives us 114, so I put that here in my answer field. Fantastic! Part B Now the second part, Part B, asks that we repeat the calculation, but now we are going to assume that 79% of adults have heard the brand. We can easily repeat the calculation with the new numbers. So I’m going to go back here and start over with my original equation for determining sample size, and now I plug in new values. We’re still 80% confident of the estimate, so our critical value doesn't change. And we still want to be within six percentage points of the true population percentage, so this value for E doesn’t change. The only thing that changes are values for p-hat and q-hat. We want to be — a survey suggests that 79% of adults of brand, so that's our proportion of success, which then means the proportion of failures is going to be the complement of that. So we’ll just subtract 0.79 from 1 to get 0.21. And now we’ve got new numbers to put into our equation. We crunch those out of our calculator, and we get 75 with a bunch of decimals behind it. So we’re just going to round up to the next nearest integer to give us 76. So I put that here my answer field. Nice work! Part C And now the last part, Part C, asks, “Given that the required sample size is relatively small, could he simply survey the adults at the nearest college?” Well, what kind of sample do you have when you sample what's nearby? I hope you said, “A convenience sample,” because that's what we've got. And of course, convenience samples are biased samples, so we don't want to be sampling with that methodology.
So we going to say, “No, you shouldn't just simply survey the adults of the nearest college because it's a convenience sample.” So we look at our answer options and find the one that corresponds with that. I check my answer. Nice work! And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to create and interpret a normal quantile plot. Here's our problem statement: Sample data for the arrival delay times in minutes of airlines flights is given below. Determine whether they appear to be from a population with a Normal distribution. Assume that this requirement is loose in the sense that the population distribution need not be exactly normal, but it must be a distribution that is roughly bell-shaped. Solution OK, here we’re going to click on this icon so we can get our data set. There's our data, and I'm going to put this into StatCrunch. Now I’ll resize this window so we can see better what's going on. Alright, now my data is here in StatCrunch.
So to check for a Normal distribution, I could just rush off to do the QQ plot, but it’s actually smarter to just do the histogram first. So I’m going to go to Graph –> Histogram, select my data, and then I want to check “Value above bar.” Here’s my histogram. Notice we have a skewed distribution — we don’t have the bell-shape — plus there's a couple of outliers here. So already it doesn't look good, but I can go make my QQ plot to confirm this. I go to Graph –> QQ Plot. I select my data. I’m going to select the box for “Normal quantiles on y-axis” because that’s how I’ve always seen the QQ plots. I hit Compute! and notice how we get a distribution of our data points here that’s making what we would call a curvilinear shape. So it’s not conforming to this straight line of best fit. And the curvilinear shape actually conforms with the skewness of the distribution which we saw earlier in the histogram. So it doesn’t look like we’re meeting the requirements for a Normal distribution. So I’m not going to select answer options B and C. Answer option A? “No, because the histogram of the data is not bell-shaped” — that’s true — “there is more than one outlier” — true — “and the points of the normal quantile plot do not lie reasonably close to a straight line.” That’s probably what we want, but let’s check out answer option D just to be safe. “No, because the histogram is bell-shaped, there’s less than two outliers” — no, that’s not going to be the one we want, so we want answer option A. Well done! And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to apply a nonstandard normal distribution to boat design. Here’s our problem statement: A boat capsized and sank in a lake. Based on an assumption of the mean weight of 132 pounds, the boat was rated to carry 50 passengers, so the load limit was 6600 pounds. After the boat sank, the assumed mean weight for similar boats was changed from 132 pounds to 171 pounds. Complete Parts A and B below. Part A OK, Part A says, “Assume that a similar boat is loaded with 50 passengers, and assume that the weights of people are normally distributed with a mean of 175.2 pounds and a standard deviation of 40.2 pounds. Find the probability that the boat is overloaded because the 50 passengers have a mean weight greater than 132 pounds.” OK, to solve this, I'm going to use the Normal calculator inside StatCrunch, and I know I need the Normal distribution because it says here that the weights are normally distributed. So inside StatCrunch, I’m going pull up my Normal Calculator by going to Stat –> Calculators –> Normal. Here's my normal calculator. I'm going to put in the distribution for the weights that are given here. So the mean is 75.2, and then the standard deviation — here it says in the problem statement is 40.2, but we've got more than one person on the boat, so therefore we have to do an adjustment to our standard deviation. And to help with that, I'm going to pull up my calculator here. So let's see — we've got the standard deviation of 40.2, and that’s divided by the square root of the sample size, divided by 50, the square root — this is the number I need to put in for the standard deviation inside the calculator in StatCrunch. I’m going to copy that. So there's my adjusted standard deviation. Now we want the probability that the 15 passengers have a mean weight greater than 132. So 132 and then this needs to be “greater than”. It looks like it's 100%, so the probability is 1. Well done! Part B Now Part B says, “The boat was later rated to carry only 15 passengers, and the low limit was changed to 2565 pounds. Find the probability that the boat was overloaded because the mean weight of the passengers is greater than 171 (so that their total weight is greater than the maximum capacity of 2565 pounds). OK, so I got the same distribution here, but I need to adjust my standard deviation. Now I’ve got 15 passengers instead of the 50, so I pull my calculator back up. So we’re going to take the standard deviation of 40.2, and this number we’re going to divide by the square root of 15. I copy that in, and I select the whole number before I copy that in — oh, excuse me, paste it in. Now I’ve got the right number. We want the probability that the passengers’ weight is greater than 171, so I need to change this number here to 171. I hit Compute!, and there's my probability — not an absolute certainty, but not that much of an improvement either. Fantastic! Part C And now, the last part of the problem asks, “Do the new ratings appear to be safe when the boat is loaded with 15 passengers?” Well, the probability the boat’s going to be overloaded is almost 2/3. That's a sizable proportion; that’s no small number! So, yeah, I wouldn’t say the boat — I mean, it’s safer but not appreciably safer. So yeah, it’s still an unsafe boat. I wouldn’t want to get on it.
So I’m going to look at my answer options here and select the one that matches that. “The ratings do not appear to be safe,” but it’s not because of this. “There is a high probably of overloading.” I like that one, but let’s check the other two before we hit Check Answer. No ... no. Well done! And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today were going to learn how to create a probability distribution table for a proportion sampling distribution. Here’s our problem statement: Three randomly selected households are surveyed. The number of peoples in the households are 3, 5, and 10. Assume that samples of size n = 2 are randomly selected with replacement from the population of 3, 5, and 10. Construct a probability distribution table that describes the sampling distribution of the proportion of odd numbers when samples of sizes n = 2 are randomly selected. Does the mean of the sample proportions equal the proportion of odd numbers in the population? Do the same proportions target the value of the population proportion? Does the sample proportion make a good estimator of the population proportion? Listed below are the nine possible samples. Part 1 OK, we see here the list of nine possible samples, and the first part asks us to construct a probability distribution table. Many students see this type of problem and it's really intimidating because they just have no clue how to proceed to solve this problem. But once you understand how to do it, it's really pretty simple. I find that this type of problem is actually easier to work in Excel than in StatCrunch. You can work it inside StatCrunch, but for this type of problem StatCrunch is really clumsy. So I'm going to work this problem in Excel because Excel does it much quicker, and it’s much easier to manipulate what we need to do in Excel than it is in StatCrunch. So here's my data in Excel. Now we’re going to enable editing. And I’m going to adjust my window here so we can actually see what is going on. OK, now we have the data here in Excel. So what we’re going to do first is look for the proportion of odd numbers in each of the samples because here the problem statement we’re looking for the sampling distribution of the proportion of odd numbers. So all we do is to look and see how many odd numbers do we have in each sample. So here I’ve got two, and there’s two numbers total, so that's 100%, which is 1. Here I’ve got again two odd numbers, so that's also 100%. Here I've got one odd number, so that's 50%. And here we’ve got two odd numbers, so that’s 100%. And I’m going to keep going down each individual row here. Alright, there's my proportions. And this border here drives me nuts, so I’m going to get rid of it here real quick. Let me blow this up so I can see what I’m doing. Whoops. Here we go. Now we’re all set. So here I got all the proportions. And now what I want to do is copy this column, and I’m going to stick it over here. We’re interested in just the numbers only. And I’m going to sort the column by going to Data and then I can go to Sort, or if you want you can just click this little quick button here because I want to sort from smallest to largest. So now we see here that we've got three numbers in our distribution. The first one is zero, so I’m going to set that here. The next one is one-half, and the last one is one. Probabilities! The probabilities are just the “part” over the “whole.” How many zeros do I have? I’ve got one, and there’s nine numbers total, so therefore my probability is the “part” over the “whole” — 1 over 9. I’m going to do the same thing for each of the other numbers. Here I’ve got four out of nine total, so 4 over 9. The same thing with the 1; there’s 4 of them over 9 total. I check my answer. Fantastic! Part 2 Now, the next part of this problem asks me to choose the correct answer below. There’s four different options here, so let’s see what we’re looking at. Here we’re comparing the proportion of odd numbers with the mean of the sample proportions. So let's go ahead and calculate that here in our cell. How many odd numbers do we have in our data set? We’ve got one, two, three, four, and I’m going to go through here and select all the odd numbers. Oh, I’m holding down the Shift key; no wonder it doesn't work. I have to hold down the Ctrl key to select individual cells. That’s why that wasn't working. And it looks like — yep, that’s all the odd numbers. Down here my count is 12 out of 18 total. So I can calculate that here; 12 over 18 is going to be two thirds. I can do the same thing with the mean of the sample proportions. Here are my sample proportions. All I have to do is take the average of that column. And it looks like they're the same. So the proportion of odd numbers in the population is equal to the mean of the sample proportions of odd numbers. So I select the answer option tells me that. Whoops! I selected the wrong one. This says “proportion of even numbers” and I want the proportion of odd odd numbers. Good job! Part 3 And now, the last part of this problem says, “Choose the correct answer below.” Again, we got four answer options here, and it looks like these answer options are talking about how good of an estimator sample proportions are. Well, sample proportions are unbiased estimators, so they're going to make a good estimator of the population proportion. And we can see an example of that here with these two numbers that we calculated from the previous part of the problem. This number represents the population, and this number represents the sample. And we can see that the two are equal, so the sample statistic is targeting the population parameter. So it makes a good estimator, so I’m going to select the answer option that tells me that. So they do target, so we don't want A or B; we want C or D. And it does make a good estimator, so you want answer option C. Nice work!
And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today were going to learn how to find percentiles using a normal distribution. Here's our problem statement: A common design requirement is that an environment must at the range of people fall between the 5th percentile for women in the 95th percentile for men. For the design and assembly of a work table, the sitting knee height must be considered, which is the distance from the bottom of the feet to the top of the knee. Males have sitting knee heights that are normally distributed with a mean of 21.7 inches and a standard deviation of 1.3 inches. Females have sitting knee heights that are normally distributed with a mean of 19.6 inches and a standard deviation of 1.2 inches. Use this information to answer the following questions. Part 1 OK, the first question asks, “What is the minimum table clearance required to satisfy the requirement of sitting 95% of men?” For this question, we really need our normal distribution calculator inside StatCrunch. So I'm going to pull up StatCrunch, and I get my normal distribution calculator by going to Stat –> Calculators –> Normal. Now I have a normal distribution calculator. The question is asking about 95% of men, so I need the distribution for the men, and that's listed here in the problem statement. So I'm going to adjust my mean and my standard deviation to match what’s there in the problem statement. We want to fit 95% of men underneath the table, so the probability of 95% is in here in this field; it’s going to show me the answer that I'm looking for, which we compute comes out to 23.8 inches. I put that here my answer field. Nice work! Part 2 OK, the second part of this problem says, “Determine if the following statement is true or false. If there is clearance for 95% of males, there will certainly be clearance for all women in the bottom 5%.” Well, that's probably true, because the bottom 5% for the women here the distribution the mean is 19.6, so that's definitely less then the 23.8 that we have here. In the 5% for the women's been a be far less than the mean, so certainly going to be yes. Although, if we wanted to prove it to ourselves, just put the distribution for the women here into our calculator inside StatCrunch and select to look for the bottom 5% that 17.6 which is less than 23.8. So yes, the statement here is true. So it has nothing to do with affect their outliers; it has to do with the 95th percentile for men being greater than the 5th percentile for women but tested. Part 3 Now, the next part of the problem asks, “The author is writing this exercise at a the table with a clearance of 23.9 inches above the floor. What percentage of men cam sit at this table?” I need to put my distribution for men back here in my calculator. The random variable here (23.9) and I'm going to get a percentage or probability coming out the other end. So it looks like we want to round to two decimal places, so that's 95.47%. Part 4 The next part of the problem asks us for the percentage of women that will fit under the table. So we do the same thing, again putting in the summary statistics for my distribution for women and press Compute! And it looks like — wow! Just about all routes to distant places again. Well done! Part 5 And now the last part of the problem asks, “Does the table appear to be made to fit almost everyone? Will the tables be fitting 95% of the men and practically all the women?” So yeah, it will benefit almost everyone. The only ones who will not fit will be about 5% of the men and a very, very small percentage of women. Nice job!
And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. Intro Howdy! I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to find a uniform distribution probability using StatCrunch. Here's our problem statement: A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.0 and 59.0 minutes. Find the probability that a given class runs between 50.25 and 50.75 minutes. Solution OK, so to do this, I open up StatCrunch. Notice there’s no icon here in the problem for you to do that with. That’s why I always love to keep a separate window with StatCrunch open and available so that I can have access to it if I need to. We need the uniform distribution calculator, so to get there I click on Stat –> Calculators –> Uniform. The next thing you need to do is put in my lower and upper limits, and those are given to me here in the problem statement — the 49 and 59. I put those in here. And then we want to find a probability in between two points here, so I go up and select the Between option. And now I put in the lower and upper limits for that section. I want to look for once it is that a first computer probability 5% but tested it and that's all there is to it.
And that's how we do it at Aspire Mountain Academy. Be sure to leave your comments below and let us know how good a job we did or how we can improve. And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you’d like to see. Thanks for watching! We’ll see you in the next video. |
AuthorFrustrated with a particular MyStatLab/MyMathLab homework problem? No worries! I'm Professor Curtis, and I'm here to help. Archives
July 2020
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