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Score on last ty: of 1 pts: See Details for more: Next question You can retry this question below Consider the sampling distribution (sample size that the cfficacy ...

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Score on last ty: of 1 pts: See Details for more: Next question You can retry this question below Consider the sampling distribution (sample size that the cfficacy of ; certain drug 0.51_ Suppose 189) for the proportion of patients cured by this arug:the mean of this distribution? What e etol of tnis distribution? What the standarddecima places Youi answers t0 RounainswcrCheck AnswerJump

Score on last ty: of 1 pts: See Details for more: Next question You can retry this question below Consider the sampling distribution (sample size that the cfficacy of ; certain drug 0.51_ Suppose 189) for the proportion of patients cured by this arug: the mean of this distribution? What e etol of tnis distribution? What the standard decima places Youi answers t0 Rouna inswcr Check Answer Jump



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Patient response to a generic drug to control pain is scored on a 5 -point scale where a 5 indicates complete relief. Historically, the distribution of scores is
$$ \begin{array}{lllll} 1 & 2 & 3 & 4 & 5 \\ 0.05 & 0.1 & 0.2 & 0.25 & 0.4 \end{array} $$ Two patients, assumed to be independent, are each scored. (a) What is the probability mass function of the total score? (b) What is the probability mass function of the average score?

So we have patients responding to a generic drug And they're going to report how well it works on a scale of 1-5 where five indicates complete relief. And the data showed that 5% I felt it was a one and 10% a two 20% of three, 25% of four and 40% of five. So this time we're gonna have to patients that are independent of each other and their scored. And we want to look for the probability mass function of the total score. So we're going to add person one with person to. So let's think in terms of an array. So I'm gonna have person one up here And they scored a one or 2 or three or 4 or five. And I'm gonna have person to down the side and they're scoring it as a one or a two as a three or a four or a five. So let's look at our totals and what they could be. So if person one score to one in person to score to one, they total a score of two. And if persons score or person one scores or to in person one scores a one they have a total score of a three. So what I'm doing is I'm just adding up and finding the possible thumbs. So when we're trying to assemble our probability mass function, I'm going to use a new variable. I'm going to call it Z could be a sum of a two or a three or a four or a five or a six or a seven or an eight Or a nine or a 10. Now to get it to there's only one way we're gonna get it too. And that is the first person scored 2-1 and then the second person scored a one. So that means the first person scored 2-1 And there was a 5% chance of that and then the second person scored a one. So we ended up with an overall probability of .0025. And then to get a three We can get a three by the first person scoring it a one And the second person scoring it a two. Or We can get a three this way where the first person scored it a two And the second person scored it as a one And that's going to yield an over overall probability of one. Now to get a four, We could have the first person scoring it as a one And the second person scoring it as a three. Or We have the first person scoring it as a two And the second person scoring it as a two. Or We have the first person scoring it as a three And the second person scoring it as a one Which is going to yield an overall probability of .03. Now we can go after a five. So if we're looking at all the fives, We could have the first person scoring it a one And the second person scoring it as a four, which there's a 25% chance of that happening. Or We can have the first person scoring it as a two And the second person scoring it as a three. Or we can have the first person scoring it as a three And the second person scoring it as a two. Or we can have the final one. The first person scores it as a four, which is a 25% chance And then the second person scoring it as a one, which is a 5% chance for an overall probability then of .065 For number six or to get a sum of a six notice there are five ways that could happen. We could have the first person scoring it a one And the second person scoring it at a five which has a 40% chance. Or The first person can score it as a two And the second person scores it as a four. Or The first person scores it as a three. And the second person scores it as a three, or The first person scores it as a four. And the second person scores it as it one, I'm sorry, as a two, Which would be .1. And then finally To get a sum of a six, we might have the first person scored as a five and the second person scored as a one And that overall probability now is going to be a .13 To get a seven. There's four ways that can happen. The first person scores it as a two and the second person scores it as a five or First person scores it as a three and the second person scores it as a four or and use this seven where the first person scored it as a four and the second person scored it as a three or for a five. First person scores it as a five and the second person scores it as a two and that gets you an overall probability of point 18. The next one Could be for a sum of eight, That the first person scored it as a three and the second person scored it as a five or The first person scored it as a four and the second person scored it as a four. That would be these two or the first person scored it as a five And the second person scored it as a three, Which is going to yield an overall probability of .22, 2, 5 to get a nine. There are only two ways that's going to happen. The first person scores it as a four and the second person scores it as a five or the reverse. The first person scores it as a five and the second person scores it as a four, Which gives you an overall% .2. And then to get a 10. That would mean both people would have to score it as a five Yielding an overall probability of .16. So therefore we could say that our probability mass function for the sum of their evaluations are of their scores or of the total score will be F of two. This .0025 F of three is .01. F four is 3. An f of 5.65 F of six is 0.13 F of seven is 18. F of eight is .22, F of nine is .2 And F of 10 is .16. So that's part A of this. Now we need to look at part B. So to look at part B, I'm going to redraw my probability distribution chances someone saying a one versus a two versus 23 versus a four and versus a five Had the probabilities of .05.1.2.25 and .4. And for part B we want to look at the average. So again, I want to create a new array because that's gonna help us organize and this time we're gonna look at the average. So person, one Might have set a one or 2 or three or 4 or five person to might have said one or two or three or four or five. And this time we're averaging their scores. So if a person once said one in person too said one, then their averages one because we add them up one plus one and we divide by the fact that there were two scores If person to said to and person one said 1, I'm going to add them up. I'll get three divide by two. So my average score is a 1.5. A three for person one and a one for person four would average out to be a two, A four for person one and a one for person to would average out to a 2.5 And then a five per person one and a one for person to would average out to be a three. So we'll keep that going for all of the rows in this array We'll have a two, 2, 5, 3, 35 and a four A four and a one average out to be a 2.5, a three, a 3.54 and a 4.5 and then a five and a one averages out to a three, A five and a two averages out to a 3.5. So now we want to look at, we'll change our variable, let's say we call it a for average we could average out to be a one or a 1.5 or a two or a 2.5 or a three, a 3.5 of four, A 45 or a five. So the probability of averaging out to a one, the only way we can get that Is if person one says 1. So there's a 5% chance that person once said one and person to says one Which is going to yield a probability of .0025. And the probability of a 1,5, There's two ways that can happen. Person one has to say a one And then person two has to say a two or person one has to say it to and person two has to say a one Yielding a probability of point out one. And then getting up to, We could have person one sale one and person to say a three or person to says to in person because one says to in person to says to or Person one says a three and person to says a one For an overall probability of .03. So I'm hoping you're seeing the pattern. If I go back up here, I had a .025, then I had a .01, then I had a .03. So my pattern is the same. So this is going to be A probability of .065. This is going to be a .13.18 .222, 5 0.2 and 16. So therefore my probability mass function for the average score will be F of one equals .0025 F of 15 is one F of two is .03 F. of 25 is 6 5 F. of three is .13 F. of 35 is 1. 8 F. Of four Is .22, 25 F. of 4 5 is 0.2, And f of five Is .16. So that concludes our finding of probability mass functions.

So we're assuming we have a normal distribution and we are given that they mean it's supposed to be 498 with a standard deviation of 100. So this would be a score of about 5 98 right here, and this would be a score of about 3 98 there. And we want to find what's the likelihood of having a score for this particular type of? I think it was an exam between 405 100. We're supposed to assume a normal population again, so we would convert both these into Z values. We take the score minus the mean yeah, divided by the standard deviation and the score minus the main, divided by the standard deviation. And that will give us our Z values and that differences negative. 98 negative night. Excuse me. Negative 98 divided by 100. So that's going to be a Z score of negative 1000.98 and this is the score is only two divided by 100. So 1000.2 and picture wise that 500 would be about here. So it's again very close to a zero Z value and the 400 is going to be about here. So we're finding that area, really? And I'm gonna use my calculator. I could use a table as well, but I'm going to use my normal CTF. And when I hit on my calculator second and distribution normal CDF, I'm going to use this as my low number negative 0.98 My upper number is going to be 0.2 and I'm leaving the mean for the standard normal distribution at zero and standard deviation of one. And when I do that, I find out that that probability is 0.34444 So I don't recall if it said percent. If it said percent, we'd get at about 34% chance of that happening or the probability is about 340.34 Then we had to look at Part B, and it said if you had a total of 300 people, how many of those would you find would have a score that's over 700? So let's find the probability of getting a score over 701st, and we can see that it's not real likely. Here's at 600 roughly and 700 would be about here. This would be at 6 98 so I can see that that probability I'll kind of color in blue is not going to be very big, which means you're not going to have a huge number here. So let's convert this into a Z score again and our Z score. What we got minus what we're assuming and think that was 4 98. It wasn't it. Yep, 4 98 and then divided by 100 and so 700 minus the 4 98. That difference is two oh two. And then we're dividing that by 100. So it's going to be two point Oh two. So that's our Z value, or about two standard deviations higher than the mean. And again I'm going to use my normal CDF here. Second and distribution go to normal CDF, and I'm going to use this as the low number 2.2 You could also look this up in a Z table and then for the upper value. I'm just going to put in 1000 and leave the mean and standard deviation at zero and one. And when I type that in. I find out that that percent is 0.217 approximately. And so now I'm going to multiply that by 300 to find out what 2.17% is of 300 times 300 and I find out that that comes out to be 6.5, so there would be probably six or seven, six or seven people I would anticipate to have a score that is greater than, uh, 700.

61. The patient recovery time from a particular surgical procedure is normally distributed with the mean of 5.3 in the centre deviation of 2.1 days. What is the Z score for a patient who takes 10 days to recover? In order to calculate our Z score, we can take our value. 10. Subtract the mean and divide by the standard deviation. We did this we get for my seven, but about 2.1, which is 2.2 option C.

The following is a solution for # seven. And this gives us a generic example where the sample Is size 15, but it comes from a normal distribution, so the population is normal. Therefore that sample size of 15 is okay. Now, had it not been normal? That sample size needs to be at least 30, but since it's normal, that sample size can be, you know, whatever, so 15 is fine, and the sample mean of this particular sample is 23.8. The sample standard deviation was 6, 3, And we're supposed to test at the significance level .01 If the mean is different than 25. Now, first off, we're going to use the T test, and the reason why we have to use a T test here is because we're given the sample standard deviation, were not given the population standard deviation sigma were only given the sample standard deviation. So, if we were given sigma, we could use the Z test, but since we're not we're given s we have to use the T test. So we're gonna do a five step hypothesis test, and the first step is to state our hypotheses and the null hypothesis h not Always assumes some sort of equality in this case, since we're testing a population mean μ equals 25. And the alternative, It says that we're just testing to see if it's different, so if it doesn't say up or down or greater than or less than, then we just assume that it's not equal to so different than 25, so not equal to 25, the 2nd and 3rd step, kind of, they kind of fall in line with each other. So I'm gonna do these together because um you can do it with the formula, but it's gonna take a little bit longer, so I'm gonna use technology. So if you have technology available, I really suggest you use that. If not, then you can start plugging in, you know, this stuff up here into your formula, but the T. Is stands for the test statistic, and then the p value is a probability. So if you go to the T. I 84 and go to stat and then air over to test, we're gonna use this T test right here, so go down, go to the T test and then make sure summary stats is highlighted. The mu not that's the hypothesized value of 25 in this case, that's just your h not The x bar, is your sample means that's 23.8 S stands for the sample standard deviation, and that was 63 And then the sample size remember was 15. And then we get down to this alternative and it's already set up as not equal to so we can just leave that the same. And then when wherever we calculate it gives us a lot of good stuff but this uh the T value and the P value are really the only things that I care about. So the T value is negative 0.74 and then the p value is 0.47 So let's go and write those down and then we'll talk about them. So the t value, the test statistic is negative zero point 738, which is pretty close to zero. So this is the so the p value is quite large, so 473 And what we do with that PVP PV I really is the most important thing um in this process. And step forward we explicitly compare the P value with our significance level alpha. And in this case it's significantly larger than alpha. P value is very much larger than our alpha value. So whenever it's larger than or whenever the PV is greater than alpha, then we fail to reject H nine. So we failed to reject the null hypothesis that μ equals 25 or more or less. We're accepting that the population mean is in fact 25 or at least we we say that it's we can't reject that. It's not or we can't say that it's not. Now. Had this been less than had the P value been less than alpha than we would reject H dot, but it's not. So we failed to reject. And then step five, we just kind of package that together into a nice conclusion and we'll say there is not sufficient evidence to suggest that the population mean, I'll just calm you for short. The population mean is different from 25. So whenever we fail to reject, that means there's not sufficient evidence to say that the whatever the alternative hypothesis is. Okay, so that's your five step T test using the P value method.


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