5

A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sampl...

Question

A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sample of 60 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.46?Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations_Answer(Enter your answer as a number

A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sample of 60 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.46? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations_ Answer (Enter your answer as a number accurate to 4 decimal places. leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 15 weeks. Assume that for the population of all unemployed individuals the length of unemployment is normally distributed where the population mean length of unemployment is 15 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 39 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is less than 14. P(X < 14) Find the probability that a sample of size n P(M 14) 39 is randomly selected with a mean less than 14. Enter your answers as numbers accurate to 4 decimal places.



Answers

Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron's, February $18,2008 ) .$ Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study.
a. Show the sampling distribution of $\overline{x},$ the sample mean average for a sample of 50 unemployed individuals.
b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean?
c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1$/ 2$ week of the population mean?

Question 27. We're told that the weekly average unemployment benefit in the US is $238 and we're also told that a researcher in Virginia state suspects that the state average there is a less than $238. So for part A were asked to develop a hypothesis test for this situation. So we can describe the Virginia state researchers view as the alternative hypothesis that Mu is less than 238 and therefore the no hypothesis is mu is greater than or equal to 238. So that's your hypothesis test for Part B. We're told that we have a sample from Virginia State of size 100 and the average of the sample is $231 and the sample standard deviation is $80 and we're asked with the T values, so we're not given the population standard deviation. So we can say that sample averages are distributed according to the T distribution, so tea is there test statistic, and that is equal to the sample average minus the population average over the sample standard deviation divided by the square root of sample size and that comes out to minus 0.88 and then were asked to calculate the P value. So just to show you what's happening here, so this is the T distribution. It's symmetrical and this is a lower tail test because the alternative hypothesis suggests that the mean is lower than the no hypothesis. So basically we have a test statistic that comes out to minus 0.88 and we're basically asking the question. Does it fall in some critical region in the lower tail such that it's low enough or extreme enough to reject the no hypothesis? So the critical region would it be defined by Alfa, which we haven't been given yet? And so the P value is the area to the left of our test statistics of the red area, so it amounts to comparing the P value to the Alfa value to see which is smaller. The P value of smaller than our test statistic is more extreme and lower in, which would lead us to rejecting the no hypothesis. But of course, the tea table gives upper tail values. So we're looking at areas in the upper tail but our test statistic if we just take the opposite of it. So plus 0.88 we would find their the area that we're looking for. So this is also equal to the P here yet, so these two tails are identical, so we can go to the table and use plus 0.88 to find RPI value. But one thing we need before we go to the table is the degrees of freedom, and that's simply and minus one, and that comes out to 99. So then, when we go to the tea table, we want to roll. That is 99 degrees of freedom, and our test statistic of 0.88 falls between these two values, which means that our P values between 0.1 and 0.2 so that's the T value. So Part C says at an Alfa equal to 0.5 what do you conclude? And we can simply say that the P value is greater than Alfa for this situation, and therefore we failed to reject the no hypothesis then for party, whereas to repeat the hypothesis test using the critical value method so we have to find her critical value. So Alfa is 0.5 and the critical value then has gotten from the table. So again it's 99 degrees of freedom. Upper tail area of 0.5 corresponds to Alfa and therefore critical values 1.660 But remember that this is a lower tail test, so it's actually minus 60.66 three. Let me just check that daily again minus 1.660 So on the left tail, critical value is equal to minus 1.66 and then to use the critical value method, we simply compare our test statistic to the critical value. And our test statistic is equal to from part B minus 0.88 and that is bigger than minus 1.66 which is the critical value. And therefore, in this case, we also feel to reject the null hypothesis

All right, so we're given data about the number of months. Certain sample is unemployed and we're also getting the number per month. So, for example, 1029 people reported they were unemployed for one month, 1686 at two months, et cetera. So first off, we're gonna make probability distribution, and I've totalled off the number of people unemployed here. That's 26,975. So for one month unemployed, we're gonna divide 1029 by 26,975. And when you do that, you'll get 0.0 381 for two months, we're going to do the same thing. 1686 divided by 26,975. That's going to be 0.6 to 5 and doing the same thing. For brevity sake, I'm not gonna explain. Every time we do this, I'm just gonna start writing them down for three. It's gonna be 0.841 for four. Zero points. 992 for 50.1293 for 60.17 to 5 for seven UM one point site. 0.1537 for 60 R for eight's my fault to certain 600.133490 point 08 62 and for 10 0.415 for part B. Need to test if what we got in part is a valid probability distribution. And to do that, we checked two things. Are all of these between zero and one? Yes, and we also need to make sure these all add up to one, and I will leave you two that I will leave that to you to verify. But this is a valid probability distribution part C. We need to find two probabilities, the 1st 1 of the probability that somebody's unemployed for two months or less. So that's gonna be the probability being unemployed for one month, plus the probability of being unemployed for two months. All right, the probability of being unemployed for 11 month excusing is 0.381 I believe. Yes, and for two months, it 0.6 to 5. You have these together you get 0.1006 now to find the probability that somebody's unemployed more than two months. While you can see that the opposite of this would be an unemployed for two months or less. So we're gonna take one and subtract probability that somebody is unemployed for two months or less. That's gonna be one minus 0.1006 which is 0.8994 That's a nine. So those are your two answers for part C party. Similar methodology. We're gonna find the probability that someone is unemployed for six months, plus the probability they're unemployed for seven months. Plus the probability that unemployed for unemployed for eight months, plus the probability that somebody's unemployed for nine months, lost the probability they're unemployed for 10 months. All right, let's look through the table to see what this So for six months, we get on retrospect. I should align this better. 0.7 are sorry. Deserve 117 to 5 plus for seven months. Zero point 1537 plus or eight months, 0.133 plus for nine months. 0.862 plus for 10 months. 0.415 If you add all these together, you'll get 0.4144 match. Her answer for party

All right. So this question gives us some information about spending on fast food and party wants us to compete the margin of error for a 95% confidence interval. So we know that margin of error equals since we only have sample standard deviation, we have to use a tea distribution. So it's t star times our sample standard deviation divided by the square root of our sample size. So we need to find our T star. So to do this, we can draw normal curve, and we know that 95% of the area it must be between negative t star and T star, so we can compete the area in each tale, which is 0.0 to 5. So to find our t star value, we just do inverse t of our area 0.975 andr degrees of freedom, which is just the sample size of minus one, which this turns out to be 1.990 So now we have everything we need to computer margin of error. We have our tea star value times air sample standard deviation divided by the square root of our sample size. And this gives us a margin of error of 1 22 point 369 Now it wants us to compete the confidence interval. So our interval is our estimate, plus or minus our error. And we know both of these. So our mean was given his 18 73 closer minus our standard margin of error from the last question. And this gives us a range of 17. 50 0.63 as are low and 1995 point 37 is our high. So we're 95% confident that the population mean lie somewhere in this interval. No, for part C, it wants us to estimate the total expenditure from this sample. So for our estimate, we estimate that a single person spends on average X bar equals 18 73 and that's in dollars. And we have 80 people in the sample, so we have $1873 per person times 80 people and that gives us a total expenditure of 1 49 a 40 for this whole group because we use our point estimate and multiply that by the number of people to get an expected total. Then Part D says that this is skewed, right? So do we expect the median to be greater or less than the mean So help? Let's draw a right skewed distribution. So this is right skewed. So for a right skewed distribution, our median would be here. But our mean would be over here because even though crest of this hill up here is the median of the data, we have these data points in the tail over here that are artificially dragging up the value of mu, so we'd expect the median to be less than the mean.

All right, were given some data about graduates and there salaries 10 years after graduation, and we've divided them into data for men and data from women. So for part A, given the statistics Ah, we're given a sample of 40 men, and we want to find the probability that our sample mean will be within 10,000 of our population. Means hold on. Gonna crack that notation. There we go on that might need his penmanship, but you get the general idea, right? So let's find our standard deviation of the sampling distribution. It's gonna be the standard deviation over sample size. So that's gonna be, uh, 40,000 divided by the square root of 40. That equals 6324 56 We're gonna find a Z lower and see upper like so. So for a C lower, uh, it's gonna be negative. 10,000 divided by our center deviation, uh, for the sample. Yeah, our sampling distribution, and then for upper, it's gonna be 10,000 positive. When you compute these out, you get negative 1.58 from 1.58 Comparing this to our normal probabilities table. This gives you probability. Lower 0.571 probability, upper 0.9429 Uh, this means our probability is gonna be probability upper minus probability, lower. Which is zero point 8858 Right party. Uh, this time we're looking at a sample of 40 women, and now we need to find the probability that our sample mean that we find is within 10,000 of the mean for the women. So once again, we're going to find our standard deviation sampling distribution. That's gonna be this time. We're looking at the statistics for the women. So this is going to be 25,000 over square root of 40. Calculate that out. That's 3952.47 are Sorry. 847 Anyway, let's find a Z lowers the upper. So the els you again. That's gonna be negative. 10,000 over our standard deviation of the sampling distribution for women. This one's gonna be 10,000 positive. So these are equal negative. 2.53 2.53 respectively. This is P. L. Looking at our table Once again, this lower probability is from zero point 0057 Upper probability 0.9943 Finding a probability, which is probability Oper minus probability, lower. You get, um, zero point 9886 part C s us two compared these to given explanation why one is higher than the other. And we see that part. He is greater. This is because standard deviation for men is greater then that for women I'm gonna have to move that up on. And because the standard deviation is greater, this means that this is a small This is smaller relative to our standard deviation, which means this let fewer standard deviations away from the mean, as opposed to this. All right, I'm gonna move to a different page party. Now we have sample of 100 men. We want to find the probability that air sample mean is within. Ah, I believe it is sorry. It's not within this time. We need to find a sample mean that is greater than our population mean minus 4000. All right, so let's find our sampling distribution. Ah, standard deviation. So that's referring back to hear. That's 40,000 square root of 100 square. 100 is tense. And this just 4000. All right, now we just see score for 4000 against 4000. So this could be because we are looking at 4000 less than the mean That's gonna be negative. 4000 top standard deviation is 4000. So this is negative 1.0 And if we look at our table, this corresponds to a probability is zero point 1587 and there you have it.


Similar Solved Questions

4 answers
3. Let A=22Orthogonally diagonalize the matrix A, by finding orthogonal matrix P (verify it is orthogonal) such that P-'AP = D, Dis a diagonal matrix.
3. Let A= 2 2 Orthogonally diagonalize the matrix A, by finding orthogonal matrix P (verify it is orthogonal) such that P-'AP = D, Dis a diagonal matrix....
5 answers
2.a) What the cause of dissohvlng = vour Nickel complex sample HCI in the determination NH;?What are the reactants vou used the synthesis of Amminenickel complex compoun d?
2.a) What the cause of dissohvlng = vour Nickel complex sample HCI in the determination NH;? What are the reactants vou used the synthesis of Amminenickel complex compoun d?...
5 answers
Andependent sampies estLevene' Test tor Equality of Variances(~test for Equaliny of Means9570 Confidence Interval of the Difference Lowci Upper 00215 55965Sig: (2 = tailed)Mean DifferenceError Difference 14055gpaEqual variances assumed0957581.99910304828090Equal variances not assumed1.96179.98505328090143260041956599
andependent sampies est Levene' Test tor Equality of Variances (~test for Equaliny of Means 9570 Confidence Interval of the Difference Lowci Upper 00215 55965 Sig: (2 = tailed) Mean Difference Error Difference 14055 gpa Equal variances assumed 095 758 1.999 103 048 28090 Equal variances not ass...
3 answers
3.10.47Queston Hclppoit and & tddar Jtalion are 5 mi apMll stralht shore funning easi and wesi A shp Iuuve s Ihe pont noan natulinlata Tatetol Z0 Inu Mtne s00 mnainlains spued ad €Qurse whal t Ihe tule 0l chanqje ol bo trackng ango 0 botwoon tho shore and Iho hno botwoon tha riklut stulion #fd Da ship 41 12 % PM? (Hint: Uso ta Nw sinosMu tuld challu out ruceonic ulu HaueheNtahmaat (Round Io loan ccinal olcu IHAulcIlu #n12 J0 PN(HcHt
3.10.47 Queston Hclp poit and & tddar Jtalion are 5 mi apMll stralht shore funning easi and wesi A shp Iuuve s Ihe pont noan natulinlata Tatetol Z0 Inu Mtne s00 mnainlains spued ad €Qurse whal t Ihe tule 0l chanqje ol bo trackng ango 0 botwoon tho shore and Iho hno botwoon tha riklut stuli...
1 answers
DranMcnikFammu nJncannaHATIsn-Kanro Apnlled untar Alre or J ZouHomework: 8-2 MyMathLab: Module Eight Problem Set roealu Scorc; 0 0'Hw Ecocoh371,41dmelCtonX5.2.9K inlttct NmkujeaFndna E # charckinla pto d dlaJaJmn rd B287 tota ##/n"tharesleota enrLa (no "ne Iunngad Ruto |Ihin clck Crick KfUXEntor your ansar nlnerExt
DranMcnik Fammu nJncanna HATIsn-Kanro Apnlled untar Alre or J Zou Homework: 8-2 MyMathLab: Module Eight Problem Set roealu Scorc; 0 0' Hw Ecocoh371,41dmel Cton X5.2.9 K inlt tct Nmkujea Fndna E # charckinla pto d dlaJaJmn rd B287 tota ##/n" tharesleota enrLa (no "ne Iunngad Ruto | Ih...
5 answers
If (X, T1) = (R, Tsorg and (Y , T2) = ([0, 2), Tu), then (X X Y, Tprod isTz but not ToT2 but not T1T1 but not T2T1
If (X, T1) = (R, Tsorg and (Y , T2) = ([0, 2), Tu), then (X X Y, Tprod is Tz but not To T2 but not T1 T1 but not T2 T1...
4 answers
Define solute, solvent, and solution by describing the process of dissolving a solid in a liquid.
Define solute, solvent, and solution by describing the process of dissolving a solid in a liquid....
5 answers
Dutcv minu Whxthv t Eari ~f converses 4ir5e s Or m con clwnon0) 26- 2 6-)n- '(6~) 0z52 (-"_ n>4) 26- n?t4 n (6f cerirsoce C) hc nC intoai 6 & # purky Sevi 4 0) 27 5) 2 2 9^ n:i 1" c) (-')n . 4) (-0)9-1 2 2 7 4) 2 n m" 2 ; 0+1 9 2 7 h) 2 2" O? n' tn
Dutcv minu Whxthv t Eari ~f converses 4ir5e s Or m con clwnon 0) 26- 2 6-)n- '(6~) 0z 5 2 (-"_ n> 4) 26- n?t4 n ( 6f cerirsoce C) hc nC intoai 6 & # purky Sevi 4 0) 27 5) 2 2 9^ n:i 1" c) (-')n . 4) (-0)9-1 2 2 7 4) 2 n m" 2 ; 0+1 9 2 7 h) 2 2" O? n' tn...
5 answers
[-/1 Points]DETAILSTIPPENS7 24.P.014. 0/6 Submissions UsedMY NOTESAsk YOUR TEACHERA 19 HC charge is 4.5 cm above an unknown charge q. The resultant electric intensity at a point cm above the 19 UC charge is 2.2 x 109 N/C and is directed upward. What are the magnitude and sign of the unknown charge? C
[-/1 Points] DETAILS TIPPENS7 24.P.014. 0/6 Submissions Used MY NOTES Ask YOUR TEACHER A 19 HC charge is 4.5 cm above an unknown charge q. The resultant electric intensity at a point cm above the 19 UC charge is 2.2 x 109 N/C and is directed upward. What are the magnitude and sign of the unknown cha...
5 answers
FAMtuVe (taftFnta nJ Kt medi ofneal MetHolr~tutar&t4 0on 35att (nTedea lethnft Menelauluntic Ueuiny Ha" IDUeisl
FAMtuVe (taftFnta nJ Kt medi ofneal Met Holr~tutar&t4 0on 35att (n Tedea lethnft Menelaulunt ic Ueuiny Ha" IDUeisl...
1 answers
In Exercises $97-102,$ use any method to solve the system of equations. $$ \left\{\begin{array}{l}{2 x_{1}+x_{2}+2 x_{3}=4} \\ {2 x_{1}+2 x_{2} \quad=5} \\ {2 x_{1}-x_{2}+6 x_{3}=2}\end{array}\right. $$
In Exercises $97-102,$ use any method to solve the system of equations. $$ \left\{\begin{array}{l}{2 x_{1}+x_{2}+2 x_{3}=4} \\ {2 x_{1}+2 x_{2} \quad=5} \\ {2 x_{1}-x_{2}+6 x_{3}=2}\end{array}\right. $$...
5 answers
98 {Graph y = tanWhat' the arnplitude and period1what* the asymptotes and its inletval'divide the asyiptole interval into equa parts find the keyvalucsBuild the completed tableand completely label Draw the gaph
98 { Graph y = tan What' the arnplitude and period 1 what* the asymptotes and its inletval' divide the asyiptole interval into equa parts find the key valucs Build the completed table and completely label Draw the gaph...
5 answers
2. The period of an element with electron configuration 1s2 2s2 2p6 352 (1 Point)3
2. The period of an element with electron configuration 1s2 2s2 2p6 352 (1 Point) 3...
5 answers
Approxlmnately Normal Wth" mean [00 and hierdistrlbution particular IQ test scores for persons over 16 years of age atendard devlationg IQ score of an SRS of 60 people Is 105 or hlgher? (Round your answer to four What thelprobablllly that the Jverage declmal Piacas )You may need to use the appropriate Appendix Table to answer this question Hla}
approxlmnately Normal Wth" mean [00 and hierdistrlbution particular IQ test scores for persons over 16 years of age atendard devlationg IQ score of an SRS of 60 people Is 105 or hlgher? (Round your answer to four What thelprobablllly that the Jverage declmal Piacas ) You may need to use the app...
5 answers
0 0 1 Veb shoes Ilsted 31 DuM IuluJJow P A1
0 0 1 Veb shoes Ilsted 3 1 DuM IuluJJow P A 1...
5 answers
6. Show that the following set of vectors forms a basis for R' (3,1,-4), (2,5,6), (1,4, 8)7. Find a basis for the subspace of R4 that is spanned by the vectors V1 = (1,1,1,1), V2 = (2,2,2,0), V3 = (0, 0, 0,3),v4= (3,3,3,4)
6. Show that the following set of vectors forms a basis for R' (3,1,-4), (2,5,6), (1,4, 8) 7. Find a basis for the subspace of R4 that is spanned by the vectors V1 = (1,1,1,1), V2 = (2,2,2,0), V3 = (0, 0, 0,3),v4= (3,3,3,4)...

-- 0.022981--