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0JpJ0ju Je survey 1 5 Round the: asks 3 final the a group of deiow 1 V Juo table declmnal lebelowy buy columnTolu E Khsao the test statistic 8 No CD expected freque...

Question

0JpJ0ju Je survey 1 5 Round the: asks 3 final the a group of deiow 1 V Juo table declmnal lebelowy buy columnTolu E Khsao the test statistic 8 No CD expected frequencies 6 1 thls students V 1 also VU decimal place} alues

0JpJ0ju Je survey 1 5 Round the: asks 3 final the a group of deiow 1 V Juo table declmnal lebelowy buy columnTolu E Khsao the test statistic 8 No CD expected frequencies 6 1 thls students V 1 also VU decimal place} alues



Answers

The table shows the number of correct answers on a test consisting of 15 questions. The table represents correct answers for a sample of the students who took the test. Find the standard deviation based on this sample.
$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Correct } & {6} & {7} & {8} & {9} & {10} & {11} & {12} & {13} & {14} & {15} \\ \hline \text { Frequency } & {2} & {1} & {3} & {3} & {5} & {8} & {8} & {5} & {4} & {1} \\ \hline\end{array}$

So a teacher has predicted, Ah, great distribution for the final exam and it was predicted that 25% will be aces, 30% will be beast, 35% will be seized and 10% will be DS. Then the exam is actually given and the great distribution is posted. And it turns out that seven people earned an A seven people earned a B five people earned sees and one person earned a d. So those would be classified as our observed frequencies. And we want to calculate the Chi Square test statistic on this so that we could run a goodness of fit test. And if we're going to run a chi square goodness of fit test and find that test statistic, we're going to need some additional information. The formula to find a test statistic is the sum of the quantity of observed minus expected squares over expected. So we now have our observed. The 775 and one are observed values. So we need to generate our expected frequencies. And if we were to add up our observed frequencies, they're going to add up to 20 students taking this test. So we expected 25% of those 20 students or we expected five A's. We would expect 30% of those 20 students to get the bees, and that's going to give us six students. We would expect 35% to get C's, and that generates seven students. And then we would expect 10% of the 20 students to get the DS, and that would be to students. Now, technically, we could not run a goodness of fit test on this because are expected, values must be greater than five. And our last category just turned out to be a two. But since we're not truly running an actual goodness of fit test on this, we're going to keep going. So we have our observed and we have are expected. So now what we need to do is we need to take each of those observed, subtract the expected, then square that value and then divided by the expected. So the first one is going to be seven, which is observed minus five. We're going to square that, and then we're going to divide that by five, and when we do that, we end up with 4/5 or 0.8, We're gonna do it for the second ones. We're gonna have seven minus six squared, divided by six. And we're going to end up with 16 and then we'll have five minus seven squared, divided by seven, which is 4/7. And I'm leaving these infraction just for the sake. They are decimals that don't clean up nice and neat. And then for the last one, we're going to have one minus two squared, divided by two, which would end up being one half. Okay, so can you see where all those values came from? And the test statistic is what we get when we add up that column. So now we wanna add up this whole column here. And when we do that, we get an answer of approximately 2.0 three 81 So if we were able to run a goodness of fit test, that is how we would find our chi squared test statistic. So this would be the answer to this problem

In Problem 82. Given the frequency distribution for the watching movies in the previous week for a survey off 25 randomly selected students. The number of movies is given here, and it's a frequency is given here for party. We want to construct history Graham off the debt this birthday instagram As here in the X axis, we put the frequency we put the number of movies in the X axis way. Put the number of movies we have zero one to three and four, and in the I axis we put the frequency. The maximum frequency is 10 or mine. We can start by zero to four. Sixth it and nine and 10, 2468 10. The first close number of movies. We have zero in the middle off the board, and it's the frequency is five. Then it's something like that. Zero is in the middle. It's a five smaller than that. Then, for one one movie, it's nine, then six, then four and finally one with that. The number of movies or in the middle off the off each part of the instagram. This is for body for, but we want to complete this table, the columns off this table, the relative frequency. We can calculate the relative frequency by dividing each frequency by the total number of students which you can, which is 25. Or we can get it by adding all these values together. Then these values sums up Toby 25. Read the frequency ISS five divided by 25 jackals point to then we have nine divided by 25.36 Then we have six divided by five, which is a 0.12 Then we have Sorry, six divided by five, 1.24 Then we have 4.1 six and finally we have four point over. By adding all these relative frequencies together, the submission should be equal to one can see that it equals year 4.6, then open toe open to its one. The cumulative relative frequency. We just calculate the relative frequency off any value and add all the values before it, which means we started about opening to then add all the values before opening toe. There is nobody's appoint toe. Then we take over 136 and the ad old values before it, which is 4.3 plus 4.2, which gives your point 56 Then we take over into and add all the briefers various to it. Or we can edit directory to the previous community breath of frequency, which gives 4.7 4.8. Then we take a 0.16 and added to appoint it. It's a 0.96 and finally we add a point offer to appoint mindsets to give one This the lost value almost the last very always equals one. Then this is the final answer off our property.

All right. So in question 36 we did a college survey asked me if they were seatbelt, and here's the response. So we're making a probability model. So the first thing we have to do is add Olive are, uh, responses. And I did that and got 4521. Then I wanna divide each category. Total 5 4521. So 118 divided by 4521 Wait 0 to 6, 249. Divided by 4521 Wait 05 behind with 076 345 divided by 4521 158 You guessed it. That would be 716 but by 4121 and 0.6 eight four. If I add those all I should get one. And the part that be, would it be unusual who randomly find a college student who never That would be this category right here. And that would be 2.6% which is less than 5%. So we consider that unusual? Yes,

So we have this data we want to make the dot plot. So the first thing we do is we take each point and we put it on the graph, and every time you put a point on the graph, just remember to cross it out, or first cross it out and then put it on the graph, which, our way, you wanna work it? But this is basically the procedure that I'm going to follow is I'm gonna take each of those points and as across each point, I'm gonna put it on this graph. So six goes right here. The next is 6.5. So it goes in the middle of six and seven and then another six. Another 79 So it really is that simple. And then we go to number four and then three and then 456 11 So 11 should be somewhere out here, so I'm just gonna put that here 6326 is So I'm gonna put the sixes here than a 10. So there's a 10 and a seven and eight. So this is how I would actually do this. Problem is, take each 0.1 by one and put a dot over the number on that number line. So here's my dot plot. So part is done now what does it say? For part? We describe the overall pattern, so I would say that the overall pattern is roughly symmetrical with one peak.


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