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4 Consider a maze of nine cells as shown in Figure 3.5. At time 0 a rat is placed cell one, and food in cell 9_ The rat stays in the current cell for one unit of ti...

Question

4 Consider a maze of nine cells as shown in Figure 3.5. At time 0 a rat is placed cell one, and food in cell 9_ The rat stays in the current cell for one unit of time Ei then chooses one of the doors in the cell at random and moves to an adjacent El Its successive moves are independent and completely uninfluenced by the food Compute the expected time required for the rat to reach the food.Figure 3.5 The nine cell maze:

4 Consider a maze of nine cells as shown in Figure 3.5. At time 0 a rat is placed cell one, and food in cell 9_ The rat stays in the current cell for one unit of time Ei then chooses one of the doors in the cell at random and moves to an adjacent El Its successive moves are independent and completely uninfluenced by the food Compute the expected time required for the rat to reach the food. Figure 3.5 The nine cell maze:



Answers

As a measure of intelligence, mice are timed when going through a maze to reach a reward of food. The time (in seconds) required for any mouse is a random variable $Y$ with a density function given by $$f(y)=\left\{\begin{array}{ll}\frac{b}{y^{2}}, & y \geq b, \\0, & \text { elsewhere. }\end{array}\right.$$ where $b$ is the minimum possible time needed to traverse the maze. a. Show that $f(y)$ has the properties of a density function. b. Find $F(y)$. c. Find $P(Y>b+c)$ for a positive constant $c$. d. If $c$ and $d$ are both positive constants such that $d>c,$ find $P(Y>b+d | Y>b+c)$.

So we're given the probably density function fifties or in your group, Its power minus 0.0 to t taken very from zero to infinity. So we need to find the property that there are two learning 50 seconds or less. So need to capture the integral 0 50 50. Did he? Okay, so is approximately point 632 and two.

Okay. The probability that rat takes 1 50 seconds or less to learn is equal 2.950213 and the property there that takes more than 1 50 seconds is one minus. Probably deal that taking 1 50 seconds or less is because the event that rat takes more than 1 50 seconds to learn is the compliment for the event that the right takes 1 30 seconds or less. So this is equal to one minus 10.950213 is equal 2.0 49787

So we're given the Providencia function fifties 0.2 Its power minus when zero duty taken very from 02 Infinity is measured in seconds. We need to find property. Then the right will learn in 1 50 seconds or less. So we need to capture the integral year to 1 50. A 50 duty, right? It was approximately 0.95 You're 2.3.

The following is a solution to number five. And we're looking at mice that go through a maze. So they let the mice mouse go through a maze and then they let him go through it again, and we're just comparing the two and we're going to look at the difference in time. So I have the data here in my tea, I 84. This is the first run that the mice get to go into 129 I think this is in seconds, that's, You know, a little over two minutes there and in the 89 seconds. And then this is the second time that they go through and it looks like, you know, some of them get better and some of them actually get a little worse. So, um that's the data. Now, I'm not going to use this, I'm actually looking at the differences because this is a matched pairs design. So I'm gonna use this L three column, go to narrow up to where L three is highlighted, and if you go to second, one minus second to we can code that column to make those differences. So you can see those differences between the first run in the second run. Okay, so now if we go to stat and then cal and one of our stats list should be L3 and then we calculate that and that gives us this X bar, that's really our D bar of 25.2 seconds and then the standard deviation sfx is actually our S and D, which is 35.661. So let's go and write those down That really answers the first two questions in all honesty. So 25 to is my statistic there and then 35 point 661 is my standard deviation. And this statistic of D. Bar that's actually my point estimate for the mean difference. So that's 25.2 as well. So it's kind of nice. Now we do the 90% confidence interval and I can use a T. I 84 on this as well if you go to stat and then test this follows a T interval because it's a matched pairs design. That's kind of a the theme of this um section. So we can go to tea interval data is gonna be highlighted. The list is L three and I'm supposed to find the 90% confidence interval. Someone can change that 2.9 and then I calculate and I get negative about 8.8 and positive 59.2. Okay, so I can write that down So the 90% confidence interval is Between negative 8.8 And 59.2 seconds. Okay, so zero is contained in that interval. So I have a feeling we're we're not going to reject but we'll see. So at the alpha, a 10% level we're supposed to test Is there a difference and more specifically do they get faster the second time around. So that's why I did greater than zero because the new 1 -7. To the new one is the first time and that should be a bigger time if they're getting faster. So this is going to give me my test statistic. And also my p value p value is really important thing because I'm going to compare that to Alpha and that's going to tell me whether to reject or not reject. So if I go back to stat harrow over to test and this time I'm just doing a regular old T. Test. It's not a two sample T test. It's really just a one sample because I'm just looking at those differences and that's just one sample. So t test and um its data and this is gonna be zero and I'm thinking that it's going to be greater than mu subzero. So then I'm going to calculate I get 158 with the p value of .09. So actually I'm gonna reject. So the T value is 1.580 And the p value is 0.095 which is just slightly under the alpha value. So I will actually reject. So I was wrong and I guess I'm going to reject the null hypothesis. So I'm saying that they are in fact getting faster their second time through the maze


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