5

Suppose that the number of dropped calls on your cell phonevaries, with an average of 2.1 calls per day. If we assume that thePoisson setting is reasonable for this...

Question

Suppose that the number of dropped calls on your cell phonevaries, with an average of 2.1 calls per day. If we assume that thePoisson setting is reasonable for this situation, we can model thedaily count of dropped calls X using the Poisson distribution withµ = 2.1.In this setting, the probabilities for 0, 1, 2, and 3 or moredropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively.Suppose that you record the number of dropped calls per day for thenext 100 days. Your observed counts of

Suppose that the number of dropped calls on your cell phone varies, with an average of 2.1 calls per day. If we assume that the Poisson setting is reasonable for this situation, we can model the daily count of dropped calls X using the Poisson distribution with µ = 2.1. In this setting, the probabilities for 0, 1, 2, and 3 or more dropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively. Suppose that you record the number of dropped calls per day for the next 100 days. Your observed counts of dropped calls are 11, 22, 28, and 39, respectively. Use a chi-square goodness of fit test to test the hypothesis that your calls are distributed according to this Poisson distribution. (calculate by hand)



Answers

When $X_{1}, X_{2}, \ldots, X_{n}$ are independent Poisson variables, each with parameter $\mu,$ and $n$ is large, the sample mean $\overline{X}$ has approximately a normal distribution with $E(\overline{X})=\mu$ and $\operatorname{Var}(\overline{X})=\mu / n$
This implies that
$Z=\frac{\overline{X}-\mu}{\sqrt{\mu / n}}$
has approximately a standard normal distribution. For testing $H_{0} : \mu=\mu_{0}$ , we can replace $\mu$ by $\mu_{0}$ in the equation for $Z$ to obtain a test statistic. This statistic is actually preferred to the large- sample statistic with denominator $S / \sqrt{n}$ (when the $X_{i}$ s are Poisson) because it is tailored explicitly to the Poisson assumption. If the number of requests for consulting received by a
certain statistician during a 5 -day work week has a Poisson distribution and the total number of
consulting requests during a 36 -week period is $160,$ does this suggest that the true average
number of weekly requests exceeds 4.0$?$ Test using $\alpha=.02 .$

So we know in a given day that they had 333 or in a given year they had 333 murders in 365 days. So that would be our mean and 333 divided by 3 65. Yeah comes up to be about 0.9123 So and I'm going to use that whole value and just store that as X. And my calculator. And we want to know the likelihood that on a given day there aren't any murders. So we know to use the poison distribution. We take the mean which will give you that 0.9123 And we're going to raise it to the X. Power. And then we're going to take E. And raise it to the negative mean power. And then we're going to take the zero which is the X. Factorial. This is one this is one. So in effect we just need to take E. To that power of negative mean and Mhm. I made a little mistake. There we go negative mean, typed it in wrong. And so I have 0.401 roughly. Six. So we have about a 40% chance of having no murders taking place on a given day. And so that's a pretty good chance that we won't have murders. But we have a better chance that they'll be at least one.

We want to run a hypothesis test based on data from World War Two in south London. So south London was divided into regions and those regions were a quarter square kilometer, and they analyzed how Maney bombs hit the regions. Some of the region's had no bombs hit. Others had 123 or four plus. And they found 229 of those quarter square kilometer regions had no bombs. HIP 211 had one 93 head to 35 had three and eight had four or more. They then ran a Poisson distribution, and they calculated the expected number of hits and found 227.5 of the regions were expected to have no bombs hit 211.4. We're expected to have 1, 97.9 for to 35 for three and 8.7 for four. Now we're going to test a claim and the claim is going to be what becomes our null hypothesis, and the claim is that the actual frequencies of hits to each region fits the calculated puts on distribution. Now, when you run a hypothesis test, you need an alternative hypothesis to fall back on. So our alternative is going to be the actual frequencies of hits to each region. Do not fit the calculated puts on distribution. And in order to run this test, we're going to have to run the goodness of fit test because we're asking, How well did the actual numbers or did the actual numbers fit the expected numbers? And in order to run the goodness of fit test, you will need to generate or calculate a Chi Square test statistic. And to calculate that Chi Square test statistic, you are going to sum up observed, minus expected values squared, divided by expected. So let's go back up to our chart, and these are your observed values. And here are your expected values. So we're going to add a column onto our chart, and we're going to call it oh, minus e quantity squared, divided by E. So we'll take the observed value minus its corresponding expected value. The answer we get will square before dividing it by the expected value. Now the fastest approach to this would be to put the data into your graphing calculator. So I'm going to just clear out a list for a second here, and then I'm gonna go stat edit. And as you can see, I have list one with all the observed values and list, too Already has the expected values. So I'm gonna sit on top of list three and I'm going to tell the calculator. Please take all the observed values from list one. Subtract their corresponding expected values from list too square that deviation and then divided by all the expected values. Enlist to and you will get these decimals. Now, for the sake of recording them, I'm going to record them out to three non zero decimals. So the first one would be 10.989 0.757 0.245 0.664 and 0.563 Now, in order to calculate that Chi Square test statistic, I must add all of the's values together again. The fastest approach would be to come back to that calculator and I'm going to tell the calculator to add up everything in list three. So, in order to do that, I'm going to quit out of there. I'm gonna hit Second stat. I'm going to scoot over to math and I'm going to hit number five to sum up list three. And in doing so, I get my chi square test statistic to be approximately 0.976 So I'm gonna come back to our screen here, and our Chi Square test statistic was point 9761534896 Now we're ready to find our P value and R P value. We're really asking what's the probability that Chi Square is greater than that test statistic? So our test statistic was 0.9761534 896 And to get a better, better handle on that, I recommend that we draw a picture. So this is what the chi square distribution would look like. It is a skewed right distribution, and the shape of it is dependent on the degrees of freedom, and the degrees of freedom are found by doing K minus one. And K represents the number of categories we have split our data into. And if we go back to our chart, you can see we've split our data into five different categories, So our K value will be five, and our degrees of freedom will be four. Now, not only does our degrees of freedom kind of dictate the shape of our curve, it also tells us what the mean of our distribution is. So the mean of our chi square distribution will also before and you find that mean always slightly to the right of the peak of that curve. So we know the four is right here. Now, we came up with a test statistic of 0.976 ish, so that's all the way back here. So our P value is our likelihood of being greater than that. So we're talking about the area off the curve into the right tail. Now, the fastest approach is to use our chi square cumulative density function on your calculator, and the calculator requires you to provide the lower boundary of the shaded area, the upper boundary of the shaded area and the degrees of freedom. So, in our problem, our lower boundary is that test statistic and the upper boundary. Keep in mind that right tail extends infinitely far. So we're going to use a very high number. We're going to say 10 to the 99th power, and then our degrees of freedom was four. So let me show you where you confined this cumulative density function. So on my calculator you're going to find it under second. There's and it's going to be number seven, number eight. So we're going to type in the test statistic, as are low boundary 10 to the 99 as our upper boundary and our degrees of freedom, which was four. And we're going to get a P value off 0.91339 now. Another component of a goodness of fit test is the chi square critical value and that is found by utilizing the chart in the back of your textbook. So you're going to go to your chi square distribution and down the left side. It asks for how maney degrees of freedom. And in our case, it's four and across the top there, asking you for your level of significance, and our level of significance is defined by Alfa. And for this problem, we want to run the hypothesis test at a 0.5 level of significance. So we're going to look across from four and down underneath 40.5 and you should locate 9.488 So we have all the components we need to make a final decision on our hypotheses. Now, when it comes time to make the decision, we could do it one of two ways. We can either use the P value that we located up above, or we could use the critical value. And again, that's an or you do not have to do both. I will show you both, and you can decide which way you like better. So for the P value, you're going to make a comparison between your level of significance and your P value. And if your Alfa your level of significance is greater than the P value, then your decision should be to reject the null hypothesis. So let's do our comparison. The Alfa on this problem is 0.5 and our P value was 0.91339 and our Alfa is definitely not greater than RPI value. So therefore, our decision is to fail to reject the null hypothesis. Now. The other approach could have been used a light utilizing the critical value, and in doing so, I like to draw a picture to represent what's going on So we would place are critical value on our chi square distribution. And in doing so, you have now separated that curve into two regions. The region in the tail is going to be your reject null hypothesis region and the region that is to the left of that Chi square is going to be the fail to reject the null hypothesis region. And you're going to then place your critical value on the distribution and our critical value. Oh, sorry. We've placed our critical value. We're gonna place our test statistic now on our curve and our test statistic was 00.976 So that would fall all the way back here in the fail to reject region. So that's why if we used the critical value to decide our decision would be the same. So either approach, we made the same decision fail to reject the null hypothesis. So let's go back and look at that null hypothesis. So we are not going to reject this. We don't have enough evidence to say that that is not true. So therefore, let's write our conclusion the conclusion to our hypothesis test is there is insufficient evidence to reject the claim that actual frequencies of hits to each region fits the calculated Hassan distribution. Now there's one last statement here or question. Does the result prove that the data conforms to the Poisson distribution and the answer? There is no. The result does not prove that the data conform to the Poisson distribution. There's just not enough evidence to throw it away. We're not necessarily showing that it is accurate, either.

So in number 86 A. It looks to define the random variable, so the random variable X would just be the length of a long distance phone call through the length of a long distance phone call. In Minutes B is ex continuous or this screen on exponential graphs. They are always continuous. See, just wants us to write the distribution. So we're told it's an exponential and we're told the mean is eight. So the parameter would be 1/8 one over the meat, so usually it's written as a decibel could be written as fraction, though that would be 0.1 to 5 for your parameter d asked for the mean the mean is eight. We were given that information in the problem f the standard deviation, the standard deviation and the mean are always equal in exponential sze. So that makes the standard deviation ain't as well f draw the graph labeled the axes, So this would be about 0.1 to 5. So you have minutes down here. Probabilities here This graph would look something like that. G. Find the probability that a phone call let last less than nine minutes, so less than nine minutes. So we're looking at all this being shaded. So since it's to the left, you would want to use one minus e to the negative 0.1 to 5 times nine. You can do that right in a T I calculator, and that gives you 0.67 53 h says last more than nine minutes. Well, the quick way would just be one minus g, and you could figure that way. But if you're looking for a formula since it shaded to the right, we would just use E to the negative 0.1 to 5 times nine, and that would be 0.3247 I asked. Find the probability that a phone call last between seven and nine minutes. So between seven and nine so e to the negative 0.1 to 5 times seven. This will find seven or Maur, but I only want 7 to 9, so I'll subtract out e to the negative 0.1 to 5 times. Not, and this will subtract out everything overnight. And that is 0.9 22 And for Jay. If 25 phone calls are made one after another on average, what would you expect the total to bay? Well, the mean was eight for each one. If we're doing 25 calls in a row, it's just eight times 25 and that equals 200 minutes.

Okay. So for this problem, we're going to test if the average phone call is greater than 2.82 different significant level Alpha one and Alpha two. So let's state the hypothesis is not is mu equal to 2.8 and each one mu is greater than 2.8. So this is our clean. So this is right to test. So we going to have the possibly critical values. So our Alfa one equal to zero poisoner one that gave us the Z one is positive. 2.33 then at Oper to that gave us the critical value. Z two IHS one Poise 65 For the test value we go into calculated by owns this formula So it would be 3.1 minus 2.8 over 0.8 over square were up 30. So this is to apoyo five. So this make the decision for our by one is you know Queen One. So at Apple one e 0.1 I have Z one is 2.33 and the test values E is 2.5 So is less than the critical value z one so it's in this region. So in this region we not reject, it's not. And then, for our two is, you know, boys, ill five. So after two is zero poison if I give us, C two is in this region, which is equal to 1.65 less than the test value is two point. Oh, by so that mean our test value is in the critical region. So in this region, we go into rejection. It's not, or we accept it want ish one, which is our claim. So the result is for upper one is 0.1 There is not enough evidence to conclude that the average phone call is greater than 2.8, and then, for the significant level is 0.5 Then we going to reject age. Not so there's enough evidence to support our claim. That means there's enough evidence to conclude that the average alcohol is 2.8 is greater than 2.8


Similar Solved Questions

5 answers
A117.9 g piece of copper (specific heat 0.380 Jlg- C) is heated and then placed into 400.0 g of water initially at 20.7*CThe water increases in temperature to 22.2'C, What is the initial temperature (in C) of the copper? (The specific heat of water is 4.184 Jlg:'C).
A117.9 g piece of copper (specific heat 0.380 Jlg- C) is heated and then placed into 400.0 g of water initially at 20.7*CThe water increases in temperature to 22.2'C, What is the initial temperature (in C) of the copper? (The specific heat of water is 4.184 Jlg:'C)....
5 answers
If at distance $r$ from a positively charged particle, electric field strength and potential are $E$ and $V$ respectively, which of the following graphs is $/$ are correct?
If at distance $r$ from a positively charged particle, electric field strength and potential are $E$ and $V$ respectively, which of the following graphs is $/$ are correct?...
1 answers
Find each indefinite integral and check the result by differentiating. $$ \int \sqrt{1+x^{4}}\left(4 x^{3}\right) d x $$
Find each indefinite integral and check the result by differentiating. $$ \int \sqrt{1+x^{4}}\left(4 x^{3}\right) d x $$...
5 answers
02 content? Valve; Chamber; Valve: Chamber: Structure (raised): Struture: 02 content?
02 content? Valve; Chamber; Valve: Chamber: Structure (raised): Struture: 02 content?...
4 answers
True (T) or false (F)?___ GAD is characterized by muscle tension, mental agitation, irritability, sleeping difficulties, and susceptibility to fatigue.
True (T) or false (F)? ___ GAD is characterized by muscle tension, mental agitation, irritability, sleeping difficulties, and susceptibility to fatigue....
5 answers
Factor each polynomial by grouping.$$m a-3 a+8 m-24$$
Factor each polynomial by grouping. $$ m a-3 a+8 m-24 $$...
5 answers
Balunced propcrly under icidic conditions_ uhabntatnc coefficients of the species shown? Whcn thc following cquationMn'-MnozWater "Ppears in the balaneed cquation #S 4(relctant, product; ncither with coclicient of(Enter for neither;)How mny clectons arC trunsfcrrcd in this rcaction?
balunced propcrly under icidic conditions_ uhabntatnc coefficients of the species shown? Whcn thc following cquation Mn'- Mnoz Water "Ppears in the balaneed cquation #S 4 (relctant, product; ncither with coclicient of (Enter for neither;) How mny clectons arC trunsfcrrcd in this rcaction?...
5 answers
T2 S= {xel:| Xu+ X2+X1=0} Fius ovlkoner_ma 8ascs th S >76 erllo_ 'fvbae proje hbb arno S tue Aar 12^ (1'0 (1 ) = ^
t 2 S= {xel:| Xu+ X2+X1=0} Fius ovlkoner_ma 8ascs th S >76 erllo_ 'fvbae proje hbb arno S tue Aar 12^ (1'0 (1 ) = ^...
5 answers
(9 pt } Tnrjnev dride mak" threx idkutical p"cn #ith I0X &rt u fnce The [xus will be" uext euch olut sharing Mc be "p' again-t Ci Tl Iwrn sik Denak Icncr- Wat dinemjot lot the total em-krure (rx tangk ineltling all URTA will muktho M01 Lag" pesile "PERS
(9 pt } Tnrjnev dride mak" threx idkutical p"cn #ith I0X &rt u fnce The [xus will be" uext euch olut sharing Mc be "p' again-t Ci Tl Iwrn sik Denak Icncr- Wat dinemjot lot the total em-krure (rx tangk ineltling all URTA will muktho M01 Lag" pesile " PERS...
5 answers
Chapter 28, Problem 048 GOA long, rigid conductor; lying along an x axis, carries current of 3.20 A in the negative X direction. A magnetic field B is present, given by B 3.26 i + 8.58 x2j , with x in meters and B in milliteslas Find (a) the component, (b) the Y- component; and (c) the Z-component of the force on the 1.01 segment of the conductor that lies between X = 1.94 m and 2.95(a) NumberUnits(b) NumberUnits(c) NumberUnits
Chapter 28, Problem 048 GO A long, rigid conductor; lying along an x axis, carries current of 3.20 A in the negative X direction. A magnetic field B is present, given by B 3.26 i + 8.58 x2j , with x in meters and B in milliteslas Find (a) the component, (b) the Y- component; and (c) the Z-component ...
5 answers
Point) A tank contains 1520 L of pure water A solution that contains 0.08 kg of sugar per liter enters a tank at the rate 3 Llmin The solution is mixed and drains from the tank at the same rate.How much sugar is in the tank initially? (kg)(b) Find the amount of sugar in the tank after minutes amount = inf(your answer should be a function of t)(c) Find the concentration of sugar in the solution in the tank after 48 minutes_ concentration 121.6 (kg/L)(kg)
point) A tank contains 1520 L of pure water A solution that contains 0.08 kg of sugar per liter enters a tank at the rate 3 Llmin The solution is mixed and drains from the tank at the same rate. How much sugar is in the tank initially? (kg) (b) Find the amount of sugar in the tank after minutes amou...
5 answers
3Let X = {xlx <-1} and Y = {xlxz-6}. Determine the given union: Express the answer in interval notation: X U Y 0 { } [-6, -1) (-oo , 30 } -6]
3 Let X = {xlx <-1} and Y = {xlxz-6}. Determine the given union: Express the answer in interval notation: X U Y 0 { } [-6, -1) (-oo , 30 } -6]...
5 answers
= ( K'x)(z' ) 4v1od 1p4" <6'h>e0 "#yiad p"wrt 7# p4 'Wail ao #tod Ivi;"' # po A Jepo1 ?L 8
= ( K'x) (z' ) 4v1od 1p4" <6'h>e0 "#yiad p"wrt 7# p4 'Wail ao #tod Ivi;"' # po A Jepo1 ?L 8...
5 answers
4) The serine in the active site of chymotrypsin functions as 0 a) a metal ion b) nucleophile c) an electrophile d) a Lewis acid )
4) The serine in the active site of chymotrypsin functions as 0 a) a metal ion b) nucleophile c) an electrophile d) a Lewis acid )...

-- 0.020206--