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A child psychologist treats four children who are afraid ofsnakes with a behavioral modification procedure called systematicdesensitization. In this procedure, chil...

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A child psychologist treats four children who are afraid ofsnakes with a behavioral modification procedure called systematicdesensitization. In this procedure, children were slowly introducedto a snake over four treatment sessions. Children rated how fearfulthey are of the snake before the first session (baseline) andfollowing each treatment session. Higher ratings indicated greaterfear. The hypothetical data are listed in the table.SessionsBaseline123477543766446677377543(a) Complete the F-tabl

A child psychologist treats four children who are afraid of snakes with a behavioral modification procedure called systematic desensitization. In this procedure, children were slowly introduced to a snake over four treatment sessions. Children rated how fearful they are of the snake before the first session (baseline) and following each treatment session. Higher ratings indicated greater fear. The hypothetical data are listed in the table. Sessions Baseline 1 2 3 4 7 7 5 4 3 7 6 6 4 4 6 6 7 7 3 7 7 5 4 3 (a) Complete the F-table. (Round your value for F to two decimal places.) Source of Variation SS df MS Fobt Between groups Between persons x Within groups (error) x Total x x (b) Compute a Bonferroni procedure and interpret the results. (Assume experimentwise alpha equal to 0.05.) Ratings of fear significantly decreased from baseline to Session 4. Ratings of fear significantly decreased from baseline to Session 3. Ratings of fear also significantly decreased from baseline to Session 4. Ratings of fear significantly decreased from baseline to Session 3. Ratings of fear also significantly decreased from Session 1 to Session 4. Ratings of fear significantly decreased from Session 1 to Session 4. Ratings of fear also significantly decreased from Session 2 to Session 4. None of the pairwise comparisons are significant.



Answers

Preliminary data analyses indicate that you can reasonably apply the z-interval procedure (Procedure 8.1 on page 330). Snakes deposit chemical trails as they travel through their habitats. These trails are often detected and recognized by lizards, which are potential prey. The ability to recognize their predators via tongue flicks can often mean life or death for lizards. Scientists from the University of Antwerp were interested in quantifying the responses of juveniles of the common lizard (Lacerta vivipara) to natural predator cues to determine whether the behavior is learned or congenital. Seventeen juvenile common lizards were exposed to the chemical cues of the viper snake. Their responses, in number of tongue flicks per 20 minutes, are presented in the following table. [SOURCE: Van Damme et al., "Responses of Naive Lizards to Predator Chemical Cues," Journal of Herpetology, Vol. 29(1) pp. 38-43 $$\begin{array}{llllll} \hline 425 & 510 & 629 & 236 & 654 & 200 \\ 276 & 501 & 811 & 332 & 424 & 674 \\ 676 & 694 & 710 & 662 & 633 & \\ \hline \end{array}$$ Find and interpret a $90 \%$ confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation of $190.0 .$

Solution number 17. And this is a, an interesting problem with the Poisson distribution. And the Poisson distribution is the type of discrete distribution where the events are random and rare. And uh, this, in this case it's a traffic accidents, Daily traffic accidents, and there is an average, that's what we use this lambda for. So λ,, which is the average of the mean is 1.72 accidents per day. Were asked to find the probability that zero accidents occur and the probability that one occurs To occur three occurs and then greater than four. Curse. Now you can use the formula, but again, I like to use the uh, software. So what I'm gonna do is I'm gonna go to second distribution and then I'm gonna go down to the present pdf and then here it asks from you, or sometimes it lasts for lambda, and that's 1.72 And then the X value, I'm just going to find the probability that zero occurs, And that gives me .1791. I'm going to go in round here some point 1791. So you can do this with any type of software or you can just use the formula. Although the formula can take awhile .1791. And I'm gonna do it one more time. Just show you whether they're the second bars for distribution. And then I went to the Plaza pdf, right? That's the probability density function. So pdf. And the mu the mean is that's the land of 1.72. And this time we're gonna find the distribution or the probability that one occurs and it's about Point Let's say .308 point 308. And that's what you're gonna do for, you know, basically all the rest of them. Until you get to the greater than so zero point 2649, I'll go ahead and give you these answers here and then 0.15 one night. So then uh to get something that's greater than what you're gonna do is you're gonna take one go all the way up to infinity and save some time. We're just going to take one minus the four that we've already found. 0123 So there's another function in the calculator we can use is called the plaza CDF. The cumulative density function. And we're going to go up to three. So the CDF calculates the probability of zero plus the probability one plus probability two plus probability of three whenever you do CDF of three, so one minus that. Or you could just do one minus these four numbers here whichever you like. So one minus. And then 2nd distribution. And I'm going to go to the present CDF. And the main remember was 172 and then the X value. Now I'm not gonna put 0123 I'm just gonna put the three And it automatically calculates 0 1, 2 and three combined. And whenever you do that that should be your answer. So .0962 0962. You might get 61 if you just use these numbers here but you get the same thing if you do one minus and these all added together. Next up we find the expected value and any time you find the expected value just take the sample size times the probabilities. So for zero, remember there were 90 days that we looked at. So 90 times the probability of zero was remember that .1791? That should give you 16.119. And then all of these are gonna be the same 90 times something. And those some things are the probabilities. So .308 that's going to give you 27.72. So that means we can expect about 27.72 days where there are there's one wreck And then 90 times the point 2649. That should give us 23 841 And then 90 times .1519. He was this 13671. And the 90 times 0.0 962 Gives you 8658. Okay, so those are the probabilities. Those are the I'm sorry those are the expected values. And then the part c we find we're gonna find our use the good as fit test. So they observed that was the chart that was given and the expected we actually just found. So we're just gonna copy those expected values down 27.72 23-841 13.671 and then eight 658. So those are the expected values where we take the probabilities times the sample sizes. Okay? So now we can go back to our calculator and do the goodness of fit test. So if you go to stat and then edit you can see that here in L. One. I put the observed values and then L. Two I put the expected values. And then if you go to stat tests it's the chi square goodness of fit test. And the observed is L. One the expected sell to or you can use the formula or any other software. And degrees of freedom was four. So the degrees of freedom, that's actually another answer. The degrees of freedom of four. Since there are five categories there of 0123 and then greater than equal to four. So we can calculate that And that gives us a chi square value of about 12 509 So chi square equals 12.509 And then it also asked for degrees of freedom. The degrees of freedom, like I said, was equal to four because it's five minus one. Okay, so the p value let's look back at the P value, it's about .01. Let's go at some .014 and that is greater than the alpha, barely, but it is greater than the alpha. So whenever the p value is greater than the alpha than we fail to reject to reject H not. Which means in this case, actually, in all these goodness of fit test cases, the null hypothesis is that the distribution does in fact follow whatever we're talking about in this case, we're failing to reject that. So this follows a person distribution so the traffic accidents, and per day in this particular area, it does in fact follow a Poisson distribution.

The key to answering this question lies in the signs and symptoms that are presented to us. So if there will be a panic in conjunction Chu in irrational fear and the panic might actually be caused by this irrational fear, we say irrational fear because there is really nothing that instigates the ah fear. Right, so panic and an irrational fear without investigation. So let's say the thing that causes fear is not present, but the fear is still there, and the panic is still there even once again when the object that is causing the fear is not present, this is going to be indicative off so together. This is indicative of panic disorder. And we simply have to remember this, uh, the way that panic disorder will present so once again, irrational fear in conjunction with panic when there is no provocation, when the object, or is the or the object of the fear doesn't have to be necessarily a real country object might be a fear of crowd places, a fear of closed on very tight spaces of fear of being alone without provocation. If the fear and panic are going to be present than this is known as panic disorder. So in the case of this question, we have a teenager who is going to have a fear of snakes and they're doing to panic while having And this is, of course, a new irrational fear. That is how it is classified. Of course, when I say well, fear of snakes is quite rational, you know, there might be scary. They are somewhere venomous, but it is still in irrational fear due to the fact that there is no clear and present danger on especially depending on where you live. There might not even be any venomous stakes nearby indefinitely in your house. There's probably not gonna be any snakes, all depending on where you live. So however, based on the question problem here, we're assuming that this is an irrational fear of snakes and the, uh, teenagers presenting with panic and has this irrational fear when this nace aren't present. So there are no snakes present there so we can conclude that this is panic disorder, meaning that choice of these the correct answer

We're going to conduct a test, an f test at the 5% significance level or non hypothesis is that the their deviations of the population are equal, and for alternative, we believe that the first one is greater than the second are calculated as statistic is going to be as one squared over as two squared, So 7.813 squared over 5.286 Squared, Which is approximately equal to 2.185. Once again, 5% significance level. So alpha is you go 2.5, No degrees of freedom are gonna be calculated as 41 miles one, so 40 and 20-1. So 19 When we go to the back of the book to calculate f of 0.05, we'll go back here, we have 40 and 19, numerator is 40 So we can go out there and then 19 oh 40 19. Here we go. Table eight. Yeah. We find out that f of 0.05 is 2.3. Nothing fancy here Now for f of 1 -0.05 or 0.95. Yeah, what we find actually, we don't even need to find it. Cool. Yeah. Okay, so this just has to do with not having to do a confidence interval at the end, been doing a couple of these problems in a row and set up is necessary. So what we're looking for here is her Critical values to three and our Have statistic is 2185. Gonna sketch this. F craft something like this. Do you point all three is right here and then the shade of region is out here. 2.185 is to the right oh my bad. 2.3. 2.03. Is that line right there? 2185 gets us to the right of that line. Therefore we will reject the null hypothesis at the 5% significance level. The interpretation is that the data provides sufficient evidence to conclude that there is less variation among final exam scores using the new teaching method.

So we are going to be performing a test for differences. Does it exist? Therefore, alternative hypothesis is as follows. Right here, we're gonna be conducting this f test at the 10% significance level. Uh for computer f statistic that's going to be s one squared over S two squared. This comes out to be approximately 1.219 Our degrees of freedom are one less than each of our sample sizes, so we're going to have 30 for the numerator and 24 for the denominator. Mhm. Now we can take this out to table eight with those degrees of freedom and we will figure out what the corresponding if statistics are. So f of mm 10% That's going to be 5% on both sides F of 0.5 You look in the back of the book here, scroll down the table eight and we find that that is 194 and now the f statistic of one minus 10.54 point 95 scrolling back down here, we see that it's 1/1 0.89 which is approximately equal to 0.53 Now is our computer deaf statistic between these two numbers 0.5 and 1.94 Yes, it is. Therefore, we fail to reject the null hypothesis at the 10% significance level.


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