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 relative risk and odds ratio for clinical trials.

Here's our problem statement: In a clinical trial of 2091 subjects treated with a certain drug, 29 reported headaches.  In a control group of 1721 subjects given a placebo, 26 reported headaches.  Denoting the proportion of headaches in the treatment group by p-sub-t and denoting the proportion of headaches in the control (or placebo) group by p-sub=c, the relative risk is p-sub-t divided by p-sub-c.  The relative risk is a measure of the strength of the effect of the drug treatment.  Another such measure is the odds ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control (or placebo) group found by evaluating p-sub-t over 1 minus p-sub-t all over p-sub-c over 1 minus p-sub-c.  The relative risk and odds ratios are commonly used in medicine and epidemiological studies. (What a tongue-twister!)  Find the relative risk and odds ratio for the headache data.  What do the results suggest about the risk of a headache from the drug treatment?

OK, so first we're asked to find the relative risk for the headache treatment.  Now we know from the problem statement that the relative risk is defined as p-sub-t over p-sub-c.  Because this is going to get a little complicated with fractions within fractions, from this point on I'm going to simply denote these two quantities as PT and PC.

So PT divided by PC is the relative risk that's defined here in the problem statement.  Now how do we find PT and PC?  Well, PT is defined here as the proportion of headaches in the treatment group.  And the proportion of headaches in the treatment group will simply be the part of that group that had headaches divided by the total number (the whole) of that group.  So 29 out of 2091 is PT.  We know this is true because of this first sentence in our problem statement: “In a clinical trial we have 2091 subjects treated with a certain drug and 29 reported headaches.”  So this is PT.  We can do the same thing to find PC.  Note that we have the part over the whole.  We’ll do the same thing to find PC.  PC is going to be 26 over 1721.  Note again we're taking the part of that group that had headaches and dividing it by the whole of that group.

A fraction within a fraction can be evaluated; we simply take the reciprocal of the fraction in the denominator and multiply it by the fraction in the numerator.  So 26 over 1721 gets flipped over to become 1721 over 26, and then we multiply that by the fraction in the numerator, which is 29 over 2091.  This gives us 49909 over 54366, which we divide out to get approximately 0.91802.  Note our problem statement says we want to round to three decimal places, so taking this number and rounding it to three decimal places gives us 0.918.  So that's the number that I'm going to put in the answer field.  Nice work!

Now we're asked to find the odds ratio.  Well, the odds ratio is found similarly by using the formula that they give you in the problem statement.  So we start by writing out that equation.  We can use the same values for PT and PC here that we used before in calculating relative risk.  So everywhere I see PT, I'm just going to substitute in 29 over 2091, and everywhere I see PC, I'm going to substitute in 26 over 1721.  This looks really awful, but if we take it piece by piece, we can actually simplify this so that it comes out really nice.

The first thing we're going to do is take care of that one 29 over 2091 that we see in the big fraction inside our numerator.  In order to subtract a whole number and a fraction, we need to first convert the whole number to a fraction.  So our common denominator will be 2091.  1 is 2091 over 2091, and then we just subtract the numerators.  So 2091 minus the 29 gives me 2062.  So now I've simplified this into a fraction.  I can do the same thing in the mega-fraction that we have on the bottom.  So again the same process — taking that one and converting it with the common denominator.  In this case it will be 1721over 1721.  1721 minus 26 gives 7 — excuse me — 1695.

So now we have fraction within a fraction within a fraction.  Gosh, this thing looks awful!  But again, we take it step by step, and we'll get to where we need to be.  Let's just move that over to the left so we have some more room to work.  In order to evaluate a fraction within a fraction, we need to take the reciprocal of what's in the bottom and multiply it by what's in the top.  But here we have layers of fractions, so we have to perform that procedure within each of the layers of our fractions.

So first, the two fractions that are on the top of the mega-fraction — I'm going to take 29 over 2091 and multiply it by the reciprocal of the fraction just beneath it.  So 2062 over 2091 flipped over becomes 2091 over 2062.  Then I multiply that by 29 over 2091.  I can do the same thing in the bottom of my mega-fraction, and you get what you see here.  Notice that we can simplify this even further.  Up here, the top part of my fraction, I've got 2091 on top and 2091 on the bottom, so they'll actually cancel out.  So when I multiply through I get 29 over 2062.  I can perform the same operation in the bottom part of my mega-fraction.  Notice 1721 on top and 1721on the bottom.  They cancel each other out leaving me with 26 over 1695.

So now we've gotten rid of a layer of our fractions, but we still have a fraction within a fraction.  So we have to apply the procedure once more.  Take the reciprocal of what's on the bottom and multiply by what's on the top.  When we do, we get 29 over 2062 times 1695 over 26.  Multiply top by top and bottom by bottom, and we get 49155 divided by 53612.  This then yields approximately 0.91687, or rounded to three decimal places 0.917.  So that’s what I'm going to put here.  Excellent!

And now the final part of our problem: What do the results suggest about the risk of a headache from the drug treatment?  Well let's review what we've actually uncovered from our calculations.  If we take the relative risk and the odds ratio and expand those out to five decimal places so we can get a better idea of what's going on, and then we also recalculate PT and PC so we have that here for our reference — now, let's look at our answer options.

Option A says, “The drug appears to pose a risk of headaches because the odds ratio is greater than 1.0.”  Well the odds ratio is not greater than 1.0.  You can see here that the odds ratio is actually less than 1, not greater than 1.  So Answer A cannot be correct.

Answer Option B says, “The drug does not appear to pose a risk of headaches because PT is slightly less than PC.”  If we compare PT and PC, we see that, yes indeed, PT is slightly less than PC.  What are these quantities representing?  Well, PT is the proportion of those in the treatment group who suffered headaches, and PC is the proportion of those in the control group — who weren't given the treatment — how many of them received headaches.  So we can see the drug has some minor effect in reducing the number of headaches.  So it doesn't appear to impose a risk of headaches, meaning there's no increase in the number of headaches, because the proportion among those who took the treatment went down.

So Answer Option B is very viable.  Let's look at the other answer options just to be sure before we select.

Answer Option C says, “The drug has no risk because the relative risk and odds ratios are almost equal.”  Well, if we look here, we see that the relative risk and the odds ratios are indeed approximately equal to each other.  However, this does not signify that the drug has no risk; the relative risk is almost a certainty —  0.918!  You've got to be kidding me that that's not saying there's no risk to it.  So that part of Answer Option C just doesn't pan out, so we can't select that as our answer.

Finally, Answer Option D says, “The drug appears to pose a risk of headaches because PT is greater than PC.  Well, again we see here PT is not greater than PC.  PT is actually less than PC.  This means we are indeed going to select Answer Option B for our answer.  Well done!

That's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below.  Let us know how good a job we did or how we can improve.  If your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find percentiles in a data set.  Here's our problem statement: Use the following cell phone airport data speeds (in megabytes per second) from a particular network.  Find the percentile corresponding to the data speed 4.1 megabytes per second.

So here we have our data set.  Let's click on this little icon to the right so we can put our data into StatCrunch.  StatCrunch will make this really easy to find the percentiles.  I'll show you how this works.  First, we're gonna resize this window for a better view of what we're doing.  OK, now that our window is resized, let me show you how to use StatCrunch to make finding percentiles really easy.

The first thing you need to do is sort the data.  Now, fortunately for us, our data set here is already sorted.  But if we didn't have it sorted, we could just come up here to Data, click on Sort and then we tell it we want to sort the data.  And then it would actually sort the data for us in a separate column which we could then use to find the percentile of interest.  Because our data is already sorted, we don't need to do that.

So the value of interest that we’re asked to find the percentile for is 4.1.  So I come over here to my list that's sorted and find the first appearance of 4.1.  Scrolling down here, I see 4.1 first appears on row 30, so I want to take the number just before it — which is 29 — and then I want to divide it by the total number in the data set.  So if I scroll down further, I can see there's 50 values in the data set.  So I want to take 29 (which is the row just before the first appearance of the value of interest, 29) divided by the total (which is 50).

So back out here on my calculator, I'm going to take 29 divided by 50 — and I can multiply by 100 to convert from decimal to percent — and I get 58.  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 care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to use the Empirical Rule to derive proportions.  Here's our problem statement: The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 246.2 and a standard deviation of 68.8 (all units are 1000 cells per microliter).  Using the Empirical Rule find each approximate percentage below.  Part A: What is the approximate percentage of women with platelet counts within one standard deviation of the mean, or between 177.4 and 315.0.  Part B: What is the approximate percentage of women with platelet counts between 108.6 and 383.8?

So for this first part of the question, we want to know the approximate percentage of the data set that is within one standard deviation of the mean.  To do that, we can just use the Empirical Rule.  Here I have a graphic imported into Excel to help us understand the Empirical Rule.  So within one standard deviation of the mean is 68% of the data.  The key to using the Empirical Rule is that you have a bell-shaped distribution, which we clearly do here in the problem statement — “bell-shaped distribution.”  So in a bell-shaped distribution, one standard deviation of the mean contains 68% of the data.  So I'm just going to put 68 here.  Fantastic!

Now, approximately what percentage of women in this group have platelet counts between 108.6 and 383.8?  Well, this would be really easy if they gave us the number of standard deviations as they did in the previous part of the problem.  But here they don't do that.  So what we're going to have to do is figure out how many standard deviations are between these two numbers that they give us here.

To do that, I'm going to go to my calculator, subtract them to find the difference, and then divide by the standard deviation, which was listed here in the problem — “standard deviation of 68.8.”  So I'm going to divide by 68.8.  That gives me 4.  So that tells me that each one of these values that's given here is 2 standard deviations from the mean, because if I go back to my Empirical Rule, notice within two standard deviations of the mean I've got one, two, three, four standard deviation units.  So we're looking at 95% of the data in between those two numbers.  So I'll put 95 here.  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 care to help you learn stats, go to aspiremountainacademy.com where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching, and we'll seeyou in the next video.

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 mean of a frequency distribution.  Here's our problem statement: Find the mean of the data summarized in the given frequency distribution.   Compare the computed mean to the actual mean of 58.2 degrees.

So we have a table here listing temperatures and frequency counts.  So let's go ahead and click this icon to the right.  We're gonna open this in StatCrunch.  And let's just move our window over here a little bit.  There!  Now that we've resized our window, we can see everything.

So we have our data here in StatCrunch, but notice these are frequency counts.  And the way we can do this really easy in StatCrunch requires us to have first midpoints for each of the classes.  What we have listed here are actually the lower and upper class limits for each of our bins, so that's not going to help us with StatCrunch.

So we have to actually calculate out here in another column what the actual frequency midpoints are going to be.  So there's my calculator.  I'm just going to take the average of my upper and lower class limits for that first class there.  So 44 plus 40 is 84 divided by 2 is 42.  So my first midpoint is 42.  I'm gonna do it again for the next class.  Punch that out — 49 plus 45 divided by 2 is 47.

Now if you notice, 47 is five more than 42, and that's because 45, the lower limit for the second class, is five more than 40, which is the lower limit for the first class.  Or you can compare the upper limits.  You can see they're also separated by five.  So I can just go ahead and just add five to each one of these midpoints here to get the rest for my table.  If you want, you could actually calculate the averages like we were doing with the first two bins, but you'll come out with these numbers here.

And now we're ready to find the mean of the frequency distribution.  StatCrunch makes this super easy!  You don't have to use that obnoxious formula that's in your textbook!  So now that we've got the midpoints and we've got the frequency counts, let's go up to Stat, I'm gonna click on Calculators, and then go down here to the bottom where it says Custom.

Values are going to be the midpoints that we just calculated.  I'm gonna select that.  Column weights are the frequency counts, so I select that column.  And then I just hit Compute! and voila!  Look here — the mean value 52.869565.  We want to the nearest tenth, so I round that to 52.9.  Well done!

And now, the second part of the problem: Which of the following best describes the relationship between the computed mean and the actual mean?  So if we look at our answer options here, the computed mean is either close or not close to the actual mean, and then the difference is being compared to five percent, so we’re either more than five percent or less than five percent.

So if I come back here to my calculator — clear that out — let's compute the actual difference so we know we're looking at.  52.9 is the calculated or computed mean, 52.8 — excuse me, 52.8 degrees listed here in the problem statement is the actual mean.  So we're gonna take the actual mean 52.8 — excuse me, 58.2 — subtract out the computed mean (52.9).  This is 5.3.  Divide by the actual 52.8, multiply by 100, and as you can see, we're at 9.1%, which is greater than that 5% threshold.  So therefore we're gonna conclude that it's not close to the actual mean because we're more than five percent off.  Check our 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 care to help you learn stats, go to aspiremountainacademy.com where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching, and we'll see you in the next video.

Howdy!  I am Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to identify and interpret critical R values.  Here's our problem statement: For a data set of chest sizes (distance around chests in inches) and weights (pounds) of 12 anesthetized bears that were measured, the linear correlation coefficient is R = 0.873.  Use the table available below to find the critical values of R based on a comparison of the linear correlation coefficient R and the critical values.  What do you conclude about a linear correlation?

OK, so here we have an icon where we can access the table of critical values for R.  So we click that.  Here's our table of critical R values.  We need to find the number of pairs of data that we have, and you're given that in the problem statement right up here where it says “12 anesthetized bears,” so 12 is the number that we want to look for.  Here's the number 12 in the table of critical R values.  Corresponding with 12 is 0.576.

Now keep in mind that you have two critical values: one positive and one negative.  Only the positive one is listed here in the table.  So we go to put our answer in.  I'm going to start by putting in this plus-or-minus sign.  You could actually list the answers separately; I think it would take it that way as well — this one negative and then comma and then the other one positive.  But I like to make it easy and not type so much.  So I'm just going to put in a plus or minus sign and then the value from the actual R table.  Well done!

Now, the next part of the problem: “Since the correlation coefficient R is” and then there's a drop down menu — “in the left tail below the negative critical value,” “in the right tail above the positive critical value,” “between the critical values.”  So in order to understand where we're at, it's helpful to draw a number line with your critical values and the value of R generated by your data.  To help us with that, I've actually produced something of a number line here in Excel.

So here we can put in our critical R values.  So we have one that's negative and one that's positive.  Now the question is “Where does our R value lie?”  0.873 lies in this region here, which is a region of acceptance.  So here in this first drop down menu in our answer fields, we're going to select “in the right tail above the positive critical value.”  And we're in a region of acceptance, so there is sufficient evidence to support the claim of a linear correlation.  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 care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy! I am Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to construct a Pareto chart from a frequency distribution table.  Here's our problem statement: A study was conducted to determine how people get jobs.  The table lists data from 400 randomly selected subjects.  Construct a Pareto chart that corresponds to the given data.  If someone would like to get a job, what seems to be the most effective approach?

So here we have our data table.  Notice we don't have the typical icon here that allows us to import the data into another place.  So we're going to have to deal with it as we see here.  We can work this problem either in Statcrunch or in Excel.  I think this problem is actually easier to work in Excel because, in order to get the Pareto chart out of StatCrunch, you're going to have to put in 42 separate counts for the first category, 52 separate counts for the second category, so on and so forth.  It's much easier to just slip these frequency counts into Excel and make your Pareto chart there.  So that's what we're going to do.

So I have Excel open here.  Let's just put in category abbreviations: the first one is H, E, N, and M, and then frequency counts of 42, 52, 274. and 32.  Now the next thing we need to do is sort the data.  It's easy to sort the data just by looking at it here because there's only four different categories.  But typically in a real world application, you're looking at something on the order of say 15-20 categories, maybe more.  And trying to organize those simply by looking at it is a tough challenge.  But programs like Excel make it super easy.

If I come up here and go to Data, and then I'm gonna click on Sort, and then I want to sort by column B.  And notice I want to sort from largest to smallest, so if you don't have that selected, come down here and make sure that that option is selected – Largest to smallest.  Let me click OK, and there's my data sorted to make a Pareto chart.

To make the actual chart, I'm going to come up here to Insert, click over here on the bar chart icon, and I just want this first option for a simple 2D column bar chart.  And here's my Pareto chart.

So now all I have to do is match the answer options with what I've constructed here and . . . N . . . M, that's the right order but the bars aren't right there.  Here we have the right order with the right bars, so answer option A is going to be the one we select.  Good job!

And now the second part of the problem.  If someone would like to get a job, what seems to be the most effective approach?  Well, that's what makes a Pareto chart so useful; it is that you can find out see right here.  There's no question.  Look at this gap here between these two bars here.  There's no question that this is what you need to be focusing on, and that's the usefulness of a Pareto chart.  It tells you where to focus your resources and your energy and your money and your time and and your manpower and whatever else you want to focus; it tells you what you need to pay attention to.

And so in this case it's going to be the N.  And N back over here in our original frequency table stands for networking.  So we're gonna select networking here from our answer options.  Check the answer.  Nice work!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below to 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 care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to identify the class width in a histogram.  Here's our problem statement: The histogram to the right represents the weights in pounds of members of a certain high school programming team.  What is the class width?  What are the approximate lower and upper class limits of the first class?

OK, so here's our data.  We have the option here to blow it up bigger if we want, but we don't really need to do that; we can see what we need to see right here.  So the class width — notice that for each of these bins (which are each of the bars that you see here), you have lower class limits listed here at the bottom of your graph.  So 110 is the lower class limit for this first bin, 130 is the lower class limit for the second bin, 150 is the lower class limit for this third bin, so on and so forth.

So the class width is just going to be the difference between successive lower class limits.  Or we could use upper class limits, but it's easier to use lower class limits because they're right here on the graph.  And it doesn't matter which two you use as long as the one that comes right after the other.

So let's just use the first two bins lower class limits 130 and 110.  So if I take 130, subtract 110, that gives me the difference, which is the class width (20).  I put that here in my answer field, and check my answer.  Good job!

Next, what are the approximate lower and upper class limits of the first class?  Well, the first class is this first bin here.  We can see 110 listed here; that's the lower class limit.  So we'll stick that there in our answer field.  And in the other answer field, we need the upper class limit.  That's going to be just barely to the next lower class limit but not quite there.  And the way we get that is by taking that lower class limit and just subtracting 1 from final digit place.

So here we have whole numbers; 130 is a whole number.  So we're gonna subtract 1 from 130 to get 129; that's our approximate upper class limit.  Check the answer.  Excellent!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below to 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 care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to construct a frequency distribution table with a specified class width. Here's our problem statement: Refer to the accompanying data set and use the 25 home voltage measurements to construct a frequency distribution with five classes.  Begin with a lower class limit of 120.8 volts and use a class width of 0.2 volt.  Does the result appear to have a normal distribution?  Why or why not?

So here we can see the answer fields forming a frequency distribution table.  The first value for the lower class limit of the first class is already filled out for us, so “begin with a lower class limit of 120.8 volts” is already done.  Now we need to take the data set and compile it into a frequency distribution table in the format that you see here.  So to do that, we're going to click on this icon so we can access our data, and then this icon here allows us to upload it into a place of our choice.  I'm going to choose StatCrunch.

So here's the data in StatCrunch.  Let's move you over here a little bit, size that to get that out of the way.  So now that we have our data set, we're done with this window; we can shut it down.  So here's our data set.

So we're going to make a frequency distribution in graphical form and then just copy the values from that graphical form over into the tabular form for our answer fields.  To do that we're going to go up here in StatCrunch and go to Graph, and I'm going to select Histogram.

First, I need to select the column where my data is located.  That's the Home Volts column, so I select that column.  Then down here under Type, I'm just gonna leave that at Frequency because we're looking for a frequency distribution; we're just looking for counts, so I leave the type alone.  Then I'm gonna put values in here in these two fields — Start at: 120.8 because we were told in the problem statement to start with 120.8 volts.  The width: We're told to use a class with 0.2 volts, so I'm going to 0.2 in here.

And then I don't know why whoever was coding StatCrunch didn't make this the default selection because it's extremely useful, but Value above bar — I'm going to check this box next to Value above bar.  And you'll see how useful that's going to be in a moment.

All the other default values are fine for our purpose, so we're going to click Compute! Oh, ho, ho, ho, ho, ho, ho!  Look at this!  Oh, it’s so beautiful.  And the best part is Value above bar gives you the values at the tops of each the bars; those are the counts in each of your bins, each of your classes.  So guess what we get to do!  Oh, we get to just transfer those numbers over!  So the first one is 2, the next one is 6, the next one is 10, the next one is 5, and the next one is 2.  So there's our frequency counts.

Now we get to put in our class limits.  We start here with 120.8.  The class width is 0.2, so the next one is going to be 121.0, or I could just look over here at my graph here in StatCrunch and I'm listed here — these are the lower class limits for each of the bins.  So I can just take those numbers and bring them over — 121.2, 121.4, 121.6.

And then the upper class limits.  Now here we were instructed — normally when you're constructing a frequency distribution, the upper limit of one bin does not match the lower limit of the next bin.  So we need the next number down from 121.0 that's not 120.8, and so that number is going to be for this particular problem halfway in between.  So that's going to be 120.9 because 120.9 gets us just to 121.0 but not exactly 121.0.

In other words, I'm taking this last decimal place, and I'm just subtracting 1 from it.  I can do the same thing here: 121.2 minus the 0.1 gives me 121.1, or I could have just added 0.2 (the class width) into this upper limit here: 120.9 plus 0.2 gives me 121.1.  And I'm just going to continue down with that same pattern to fill in the rest of the table — [121.3], 121.5 and then this one's going to be 121.7.  Check my answer.  Good job!

So I click Continue to get the rest of the problem.  Does the result appear to have a normal distribution?  Why or why not?  Well, looking back here in my graph in StatCrunch, I can see that the approximate pattern for a normal distribution is for the values to start low, come up to a high point in about the middle of the range, and then come back down to a low value again.  That's the pattern that we see here, and so therefore, I'm going to say, “Yes, the frequencies start low, reach a maximum, then become low again, and are roughly symmetric about the maximum frequency.”  Notice the word rough.  Roughly is exactly what it means — roughly. It's not exact, so here we go.  We're gonna select that answer option.  Good job!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below to 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 care to help you learn stats, go to aspiremountainacademy.com where you can find out 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.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today, we're going to learn how to calculate and compare an exact value with a practical value.  Here's our problem statement: A polling company reported that 19% of 1,018 surveyed adults said that pesticides are “very harmful.”  Complete Parts A through D below.

What is the exact value that is 19% of 1,018?

Well, here in this case, to solve this question, we're going to translate the English into math.  So we're looking for the exact value, so we could call that X or whatever you want to call it.  Is translates into an equal sign.  19% — we can convert that to decimal form so we can perform a mathematical calculation.  Of translates into multiplication.  And 1018 is a number we're going to multiply by.

So I'm just going to go over here to my calculator, and we're gonna multiply 19% — just 0.19 (just move that decimal point over two places to the left to convert from a percent to a decimal) — of is times 1018.  That gives us 193.42, so that's what I'm gonna put here in my answer field.  Good job!  Next part . . . .

Could the result from Part A be the actual number of adults that have said that pesticides are “very harmful”?  Why or why not?

Well, if you want the actual number of adults, you're talking about discrete data because we don't count partial people.  We count people as whole values, so the actual number of adults is going to be an integer value.  The exact value 193.42 is not a whole number; it is a decimal.  And so this represents what we call continuous data.

So the answer to the question here in Part B is going to be “No” because we're only counting whole people.  So right off the bat we can eliminate answer options A and D.  So let's look at answer options B and C.

B says, “No, the result from Part A could not be the actual number of adults who said that pesticides are very harmful because that is a very rare opinion.”  No, we're actually talking about the difference between discrete and continuous data, so that doesn't sound right.

Let's look at answer option C: “No, the result from Part A could not be the actual number of adults who said that pesticides are very harmful because a count of people must result in a whole number.”  Bingo!  That's what we're looking for — “must result in a whole number” so that gives us discrete data.  We check our answer.  Good job!

What could be the actual number of adults who said that pesticides are “very harmful”?

Well, to get the actual number, we're just gonna look at the exact value that we calculated here, and we're going to round it.  So 193.42 rounded to the nearest whole number — we're gonna go down because the number right after our decimal point is less than five.  So that gives us 193, which we type here in our answer field.  Well done!  And now the last part . . . .

Among the 1018 respondents, 373 said that pesticides are “not at all harmful” — I’m not sure what they're . . . they're probably like, you know, making some sort of drug out of the pesticide and snorting it or something ‘cause I don't know how you can say pesticides are not at all harmful.  It's a . . . it's a . . . it's a processed chemical, man!  It's gonna . . . it's gonna have some harm to it!  But anyway that — that's what they said.  We’re not here to judge; we’re just here to take the data and do the stats with it.  So 373 said that they were not at all harmful.  What percentage of respondents said that pesticides are not at all harmful?

So if we want a percentage from the raw data, we just take the part and divide by the whole.  So the part is the 373; that's the part of the whole 1018 respondents in the group.  So I'm just gonna come over here, clear out my calculator, take 373 (that's the part) divided by the whole (1018), and boom!  There's our answer!

Now this is the percentage in decimal form.  Notice that here in the answer field, we're asked to calculate a percent; it wants the answer in percent form.  So I can move this decimal point in my head and put this in, but it says rounded to two decimal places.  So just to make sure I put the right number in, I'm going to come in and multiply this number by 100.  So then I convert from decimal to percent.  So now here's my percent which I can round the two decimal places — 36.64 — so that's what I'll stick in here.  Good job!

And there it is. That's the solution to each step of the problem.  That's how we do it here at Aspire Mountain Academy.  Be sure to leave your comments below to 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 care to help you learn stats, go to aspiremountainacademy.com where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching, and we'll see you in the next video.