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4at MMT -Do Hurtal' SaelYalcninonEAtonnictHusdo Makecilld 47dx Dhack DoAssinnm Esorthnn Yurt Id 57045equestionid https / /www mathxlconStudent PlayetHomeworkShawn Rumery4/3/19 8.28 PMMAT-240-R4676 Applied Statistics 19EW4 Save Homework: 2-2 MyStatLab: Module Two Problem Set HW Score: 52.22%, 33.95 of 65 pts 13 of 15 (12 complete Score: 0 of 5 pts Question Help 8.1.29-T commercials Complete parts (a) througn monitored by firms because Ihis helps to determine advertising pricing for The amo

4at MMT - Do Hurtal' Sael Yalcninon E Atonnict Husdo Makecilld 47dx Dhack DoAssinnm Esorthnn Yurt Id 57045equestionid https / /www mathxlconStudent PlayetHomework Shawn Rumery 4/3/19 8.28 PM MAT-240-R4676 Applied Statistics 19EW4 Save Homework: 2-2 MyStatLab: Module Two Problem Set HW Score: 52.22%, 33.95 of 65 pts 13 of 15 (12 complete Score: 0 of 5 pts Question Help 8.1.29-T commercials Complete parts (a) througn monitored by firms because Ihis helps to determine advertising pricing for The amount of time aduits spend watching television is closely (d) Let 0a (Round to six decimal places as needed: ) Keexday of pelween and hours adults results in mean time walcning television on probability tnat random sample 0f 60 (c) Determine (Round t0 iour decimal Jaces &s needed The probability is 0.9513 random sample of 55 individuals who consider Inemselves reduce television watching: Suppose that saniple mean of 2.11 nours less from of the Internet i5 that it is thougntato GAkday Deterinine tne likelinood of obtaining (d) One consequence 0f tne popularity in a mean time of = nours walching television On t0 be avid Internet users rosuils population wnose mean presured to 1e 2.45 hours (Round t0 four decimal laces as needed: / The likelinood answer box and then click Check Answal Enter your answer in tha parts 2 pemaining Type here [0 search



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Preliminary data analyses indicate that you can reasonably apply the $t$ -interval procedure (Procedure 8.2 on page 346 ). According to Communications Industry Forecast, published by Veronis Suhler Stevenson of New York, NY, the average person watched 4.55 hours of television per day in $2005 .$ A random sample of 20 people gave the following number of hours of television watched per day for last year. $$\begin{array}{ccccc} \hline 1.0 & 4.6 & 5.4 & 3.7 & 5.2 \\ 1.7 & 6.1 & 1.9 & 7.6 & 9.1 \\ 6.9 & 5.5 & 9.0 & 3.9 & 2.5 \\ 2.4 & 4.7 & 4.1 & 3.7 & 6.2 \\ \hline \end{array}$$ a. Find a $90 \%$ confidence interval for the amount of television watched per day last year by the average person. (Note: $\bar{x}=4.760 \mathrm{hr}$ and $s=2.297 \mathrm{hr} .$ b. Interpret your answer from part (a).

So we would be assuming that the mean number of the minutes for Disney movies and the mean number of minutes of non Disney or other movies, Children's movies showing tobacco, that those two are equal and alternately that they're just different. So again, we're looking at how many minutes or how many seconds actually we're seeing tobacco shown. So we're going to assume that the difference between these two groups is zero and we're doing a two tailed test. And so let's find what our test statistics are. Test statistics is not going to be and are smaller sample sizes 17, So we'll say we have 16 degrees of freedom. And so I take the difference 61.6 seconds for Disney minus 49.3 seconds, or the other Children movies. Okay. And then we have very, very large standard deviations. And that is something to really pay attention to notice that those standard deviations look how much larger they are than the actual even mean. And so normally, if you look at these two times you're gonna think, well they're going to be significantly different because, you know, they're over 10 seconds apart. However, with those huge standard deviations that's going to make that be suspicious, and when we get that we get 0.46 So we're getting a test statistic here, that is a T. Value, that is 0.46 and this area plus going symmetrically on the other side, those two together are going to be our P. Value. So we need to find two times the probability of that being greater than or equal 2.46 And I used my T. C. D. F. On my calculator and found out that that p value is about 65%. So if you have two distributions that their difference is actually zero, the likelihood of getting what we got, or more extreme in each direction happened 65% of the time, and this is definitely not less than 5%. Therefore we have we failed to reject now. Yeah. Okay. So it appears as though the means are equal, so they mean amount of time that you see tobacco means are equal even and again, because that's standard deviation is so large, That's what's causing that to happen now. Since we did a two tailed test, we would be finding a 95% confidence interval of our significance level was 5%. And so I can look up the 95% number for 16 degrees of freedom and for 16 degrees of freedom, I get that T. Star value being 2.1 to 0, so we can take that difference. And this difference looks like that is 61.6 minus the 49.3 comes out to be 12.3 plus or minus R 2.12 zero. And then the square root up. And let's get those big standard deviations here square divided by the sample size of 33 and that's 69.3 square divided by the sample size of 17. So let's find that margin of air 2.12 point 12 times the square root of 118.8 squared, divided by the sample size of 33 plus the 69.3 squared divided by the sample size of 17. And I get a margin of air of a whopping 56 basically 560.5, which I'm going to store that as X. And then we're going to take that 12.3 and subtract away X. And I get my low value as a negative value, negative 44 point roughly. Two. And then when I changed that and I add those values together, I get the upper limit as 68.8 and notice that it does include zero. So it includes zero where 95% confident that the difference between the two means is somewhere in here and it includes zero. So we would not have evidence to reject the not we would fail to reject it. And uh and I think you had a part, see that stated something about looking at those values and when you look at those, it appears to me that there are lots of values that are less than uh 25. About half of the values are less than 25. Less than 25. Almost half of them. Therefore you are going to have skewed data so it won't be normal. So we would have to be be careful be careful about using these results because we are assuming that the two distributions are normal and the test is pretty robust. So, um, so, but you know, we if we had some more samples, that would be better for us.

All right, we're looking at the number of hours watched by some students, and that's what this data is, and I've sorted it uh into a column. And so I can easily do some calculations with it. And they were asked to figure out or determine if this normal probability plot which is created from the data, is Yeah, it tells us uh that could be normal, and the answer is yes, because the way these these probability plots work is that if you have your point, your data points within these bounds, you can assume it's normal. There's a really good chance is from a normal distribution. And so according to the normal probability plot, which I recreated here, Yes, we're told to find the sample mean and sample standard deviation. And so I found those here, using the spreadsheet function average and standard deviation. So that's what those are. And then using part B estimates, we are told to graph the role model for the distribution. So, what I did here, and I'll do that to and did that twice. Why do it once in a spreadsheet, when you can do it twice for the main standardization, I just want to draw a normal model for the distribution. So we'll go ahead and do that. And that is part C. So, here's my normal curve in the mean, is 20 point four, which is the mean. And then we're gonna add a standard deviation to get the next answer. Here's uh huh who is that? And that is going to be this is plus, I'm just going around this to 10 point five Most 10.5. That's gonna give us, what's that going to be 30 14 for the first standing ovation and then we'll had another one up there and that's gonna give us 40 yeah 41 point 31 point nine. Um And then you do it again That and that's another plus 10.5. And that's gonna give us, What's that going to be? 51.4 I believe when he's double check. 52 please. 52 52.4 Okay. And then we've got let me do that subtract. Okay, let me do keep someone consistent here -10. It's going this way and there's Is that? And that's gonna be 10 that's five. I'm just gonna keep that before I'm just going around at the 20.9 because I'm going to keep track of the thousands of hours. I watched tv Mhm. There's the second Last another 10 so that's going to be 10 points. Oh that is 10, whoops, I didn't it's color so it's gonna be negative. So this is a it's negative. Uh Mhm 0.1 Mhm. 12 Yeah 10 for you for this 10. Mhm Okay. You know what, I don't know why my brain sometimes our brains after a long day of a hard time calculating. So there we get into negative values. And uh I was talking to a friend of mine who just who had to statistics with back in college, That's a negative 10 point, right? And uh you know sometimes statistics just doesn't give the values that don't make sense. You have to be okay with it because you can't have negative time. I mean that doesn't work. But still that's our values. These are normal drawing. And then we want to figure out the party the probability that a student has watched between 20 and 35 hours. So for that I'm going to use the spreadsheets again. So this is D. D. And that's uh Z scores. And if you're called Z scores Z is X minus mu over sigma. And I'm gonna use a spreadsheet for this. So we need to find the Z score of 20 and the Z score of 35. And then use the normal probability distribution table. Or maybe even the a computer program to calculate it. So to do the area. So that's what we're going to do. So let's get a picture. I was like pictures, pictures are nice. So here's the here's the curve And we want to know between 20 which is a little less than the mean. So here's the mean. Didn't black, Here's the mean. And then we want to know between 20 Which is roughly here in 30, which is up here, we don't know this stuff. And so what we need to do is find the Z score and then the area under the curve which will give us this red area, and then Well, look up the Z score for this, this is the 3rd, 35, and then we'll get this green area. So what we do then, is to do the green like this, and then we're gonna subtract off the red, and that will give us, I was like equations with pictures, these are kind of fun, something like this. God. There you go. That's what we're looking for. This is the this purple pinkish era we want. So I did that for 20, and actually I just use this normal distribution function in the spreadsheets, which you put in the X. Bar or the the sample point, the mean, which we calculated the standard deviation, which we also calculated, and then it gives us the appropriate air into the curve, as opposed to doing the Z score. And then the look up. So let's do that for both of those were the difference, we subtract them. So the probability uh the probability is point um This .44, really? And then e the last part we want to use the model that we created earlier, Uh to see if the probability of picking a student who watched more than 40. Oh, Hey, that's a lot. So if we use this idea of the picture, it's gonna be hears this and 40s up here. So more than 40s over here. So how do we get that? Well, up to 40, is this green stuff. So to get the black, what we do is we take the whole curve which I'll cover in red with the whole curve. Mhm. Excuse my drawing The whole this is the whole curve and then we're gonna subtract off the green and then we'll be left with our area that we want. Which is this little black Nevin over here. So to do that uh we do this I do the same thing with the normal distribution function in the spreadsheet. Put in the appropriate mean and discrimination. And then That gives us this number .965 is the green area here. And then to calculate the area that we want is one minus that value which is 10.3. So the probability of picking a student Who watches more than 40 hours a week as a point of three. Okay. Probably that Bill t. There you go.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

So I'm gonna let d stand for just for convenience. The beginning time minus what the end time is. And we can subtract the opposite way as well. But it talks about is there a difference? So we would assume that that difference Is non existent, that there is no difference between the beginning and the end so that the main difference is zero and alternately that it's different. So they didn't say that they wanted to test to see if the time had gotten smaller. They just asked if it's different. So We have a total of eight people And we're doing a two tailed test and we want our significance level to be 1%. So we need to split that 1% between the two tails. So this needs to be 20.5 and this needs to be 0.5 And we'll find that T. Value with eight degrees excuse me, seven degrees of freedom. That T. Star value with 5% of 50.5% in that upper tail and that corresponds to with 7° of freedom. Looking like that is 3.499 and this one is negative 3.499. So we would reject the null Any time we have our test statistic which is a t. value with 7° of freedom being less than negative 3.499. Mhm. Or if that test statistic is greater than 3.499. So we can actually plan all this out before you actually look at the data and that was basically what we're doing. So part C. I subtracted all those values and hopefully I didn't mess up, I just did them in my head And I got the difference to be 8.2375 and the standard deviation of those values It was 5.2- two and it would round off to five. Now we want to find the test statistic and I have these values started my calculator And so our test statistic was 7° of freedom is going to be that mean that we got minus the mean we're assuming which is zero and then divided by the standard deviation Over the square root of and make sure you use eight and not seven. And when I did that I got my test statistic to be 4.461. Now look where that lies, that's going to lie up here and I can see that is in the rejection zone. Therefore part E we have evidence to reject the null. Mhm. Yeah. And that means that we would conclude that there is a difference, There is one is a difference in the amount of time. It doesn't ask us to say what direction, but it surely looks like the before is higher than the after.


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