5

The mean and standard deviation of random sample of n measurements are equa t0 33. and 3.1, respectively:Find 99% confidence interval for H ifn= 100 Find 99% confid...

Question

The mean and standard deviation of random sample of n measurements are equa t0 33. and 3.1, respectively:Find 99% confidence interval for H ifn= 100 Find 99% confidence interval for p ifn= 400_ Find the widths of the confidence intervals found in parts and sample size while holding the confidence coefficient fixed?Wha: the effect on the width of coniidence interval of quadrupling theThe 99% confidence interval for | ifn=100 is approximately (Round t0 three decima places as needed:)The 99% confid

The mean and standard deviation of random sample of n measurements are equa t0 33. and 3.1, respectively: Find 99% confidence interval for H ifn= 100 Find 99% confidence interval for p ifn= 400_ Find the widths of the confidence intervals found in parts and sample size while holding the confidence coefficient fixed? Wha: the effect on the width of coniidence interval of quadrupling the The 99% confidence interval for | ifn=100 is approximately (Round t0 three decima places as needed:) The 99% confidence interval H ifn=400 is approximately (Round t0 three decimal places as needed ) Choose the correct answer below Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the coniidence interval by factor 0f 2 Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the confidence interva factor 0i 4. Quadrupling the sample size while holding the confidence coefficient fixed does not affect the width of the confidence interval Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the coniidence interval by factor 0f 4_ Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the confidence interval by factor 0f 2



Answers

A simple random sample of size $n$ is drawn from a population that is normally distributed with population standard deviation, $\sigma,$ known to be $13 .$ The sample mean, $\bar{x},$ is found to be 108. (a) Compute the $96 \%$ confidence interval about $\mu$ if the sample size, $n,$ is 25 (b) Compute the $96 \%$ confidence interval about $\mu$ if the sample size, $n$, is $10 .$ How does decreasing the sample size affect the margin of error, $E ?$ (c) Compute the $88 \%$ confidence interval about $\mu$ if the sample size, $n,$ is $25 .$ Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the size of the margin of error, $E ?$ (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why? (e) Suppose an analysis of the sample data revealed three outliers greater than the mean. How would this affect the confidence interval?

Following a solution to number 12, and this gives a summary stats that the sample main X bar is 35.1 and the sample standard deviation was a 8.7. Now notice it's a sample the esses sample standard deviation, and we don't know what sigma is, the population standard deviation. So since we don't know sigma, we have to use the tea interval instead of the z interval. Remember the less, you know, the more variable is going to be. So the t interval is what we're gonna use here, since we know s not sigma, and we're asked to find three things and then we talk about Um a little bit of conditions for inference on this last part, but let's look at a B&C because they kind of go together first, we're gonna find the 90% confidence interval winter and is 40. Then we're going to find the same level of confidence, 90% confidence interval, but this time the end is 100. So we increase that sample size And then this 3rd part We increase the confidence of them, but we keep the N equal to 40 again. And then we're going to talk a little bit about the conditions for inference. Now in the beginning it did not say anything about The population distribution. So normally it says assume that the population distribution is approximately normal. Well, here we don't need it to be normal because N. Is large enough. That magic number is 30. As long as your sample size is at least 30. It doesn't matter how your population is distributed. Um Because that's just the role as the central limit theorem world. Alright, so I'm gonna use technology for this because it cuts down. But you can use any excel. I'm gonna use the T. I. T. For but you can use any sort of technology Wish. And if you'd like to use the formula you should get the same answers. But I'm gonna use T. I. D. Four. So if you go to stat and then air over the tests it's this eighth option here the T. Interval. So you can scroll down or you can just click eight and make sure the summary stats is highlighted. So we don't have any data. Um we don't have a long list of data or anything. We're just giving the summary stats. So keep that highlighted and it asks for X. Bar. That's 35.1. S. Was 8.7 in was 40 at least for this first one and then the sea levels 90% of 900.9 For this first one. And we're gonna go and calculating this first band here. That's our interval. So 32.8 To 37.4 is r. t. interval. So 37 point I'm sorry 30-32 point 782 To 37.418. Okay so that's kind of like our base. So it's about uh we can kind of estimate maybe about five a part of 4 4.7 units apart. So just kind of keep that in the back of your mind 4.7 units apart. So let's see what happens whenever we increase that sample size to 100 and you may be able to guess by now. So if we go to the T. Interval again this time I'm just going to change this into 100 And let's see what happens. So we go from 36 I'm sorry 33.6-36.5. So let's write that down. Let's compare 33 .655. All the way up to 36 545 Okay. So notice that the lower bound gets higher and the upper bound gets lower. That means remember appear we were up to 4.7 here. It's You know about three actually a little bit less than 3-9. That interval um is Only 2.9 units so it's gotten narrower. So the next part of this question is how does the sample size or how does how do the degrees of freedom influence that margin of error? Or intern how does it influence the interval with? So as in increases or as the degrees of freedom increase, that margin of error is going to decrease. Which makes so making the confidence interval more narrow. Okay, so as you increase your in and you can kind of think about that standard air that? S over square devin, you're dividing by a bigger number which makes that margin of error smaller um and the smaller the margin of air then more narrow your confidence interval will be. So now let's look at the 98% confidence interval. But this time the ends back to 40. So if we go to stat Tests and it's that 8th 1 And this time we're going to bring that down to 40 and then we're gonna bring this one up to 98% of .98. We calculate and we get 31- 38. 31.8- 38.4. So it's right that down 31 point 763 two, points 437. So we're gonna compare this one to part a where it has the same sample size, just different confidence level. So we increase the confidence level and notice that we went from 4.7 units apart. And this is like 6.7 units apart. So it's gotten wider. All right. So What's that relationship? Well, as the confidence level increases, which is what we did here went from 90 to 98. That margin of air also increases because that intervals wider. So making the confidence interval wider. So the more confident you are, the wider and wider that interval must be. And then part D. It says what conditions of inference are needed if N equals 18. So if N equals 18 and I kind of sort already hinted on this, But since that end is less than 30, the population must be normally distributed since And is less than 30. Okay, so since N is less than 30, That population needs to be normal. Now. Normally they say that in the beginning but this time they didn't because these sample sizes are big enough, it doesn't matter. But if that in were equal to 18, 1 of those conditions for inference, if it is less than 30 is that the population comes from a normal distribution.

Problem 22 or with the data values from the smallest to the largest, which is hundreds, hundreds on one 1/6. Then 115 115 141 one or two and one for three and 14 green and one for five and one for seven and one for seven and 1 15 and 1 52 on 1. 53 and 1 55 and 1 57 and 1 61 63 and 1 64. So to remind the critical value using Table and J, who's offer is equal to opening one and then is equal to 20 so the critical value is equal to find, so add one toe critical. Various okay is equal to five plus one, which is six to the boundaries, off the confidence interval for the million or then six smallest and the sixth largest value insulting data list, which is 1 41 and 1 53

The following is a solution to number 11 and were given summary stats that the sample mean X bar is 18.4 and the sample standard deviation is 4.5. Now we only know the sample standard deviation, not the population standard deviation. So since we don't know the population standard deviation sigma, we have to use the tea interval instead of the z interval, so keep that in mind moving forward. So I'm gonna use technology for this problem here. We basically have three questions that are very, very similar, and then um one where we talked a little bit about conditions for inference, um now it doesn't say anything about the normality of this population, that's going to come up here in part D, But it doesn't matter because these sample sizes are big, they're greater than 30, so it doesn't matter how this population is distributed, that sample size is big enough. So we're asked to find the 95% confidence interval whenever end is 35, Then we find the 95% confidence interval will never end this 50, we're gonna compare that so the same confidence level but the sample size is a little bit bigger. Then we come down here and we increase that sample that confidence level of 99% confidence. And then we're back down to the n equals 35. So let's take a look and see what happens. So I'm gonna use the T. I. T. Four but you can use any sort of technology want or you can go and do the formula but if you go to stat and then arrow over two tests and it's this eighth option here, the T. Interval and um make sure that the summary stats is highlighted because we don't have data. We're just giving summary stats and X bar was 18.4. The s was 4.5. The end for this first one was 35 then that confidence level is 350.95 and that represents 95%. So whenever we calculate this top band here that's our confidence level. So about 16.9-19.9 is going to be our interval. So let's go and write that down. So 16 Point I'll go ahead and do what they did. So 8854 And then the upper bound is 19.946. Okay So that's our first confidence interval. Now let's see what happens when we increase that sample size from 35 to 50. So we go back to stat tests and it's this eight option here, the t interval And this time the only thing that changes is the end and that's 50 now. So we calculate And we go from about 17 to 19.7. Okay, so let's write that down. So 17 .1-1 All the way up to 19 six 79 So let's compare now. So it looks like the lower bound has actually increased by, I don't know maybe about point 2.5 4.3 and the upper bound has decreased. That means that interval is narrower. So what does that say about in? And the margin of air? Well, as in increases, which is what happened here? The margin of air decreases. Making the confidence interval more narrow. Okay, so they're inversely related. So as in increases or degrees of freedom, increase. That margin of error is going to decrease, making it less variable. And that's the reason why that interval is a little bit narrower. So now let's go back to the 35 sample size. But this time we're going to increase that confidence level to 99%. So go back to stat tests and the 8th option. Okay. And we're back down to 35 but we're going to increase that sea level of the .99. And then when we calculate, let's see what happens here. So we go from 16.3 to 20.5. Okay, so let's write that down 16 point 3 to 5, 20 point 475 So compare this to part A where we had that same sample size but different confidence levels. So the lower bound actually decreased in the upper bound increased. That means that interval is wider. Okay, So what does that mean as the confidence level increases? That margin of error is also going to increase because that that critical values actually greater. So this is larger whenever you increase that confidence level, um making the confidence interval wider. So as the confidence level increases, that confidence interval is going to increase or get wider as well. All right. And then part D. Now it says what conditions for inference are needed? If in equals 15? Well, if N equals 15, then the population from which the sample was taken must be normally distributed. Okay, Since And is less than 30. Okay. So notice in part A B and C and I already talked about this, but A B and C. Um those sample sizes are greater than equal to 30, so it doesn't matter how this population is distributed. It still works. We can still use that procedure But if N is less than 30 in this case 15, then we need that statement in there that, Hey, don't worry about the population, it's normally distributed.

The following is a solution to number nine and this were given the sample mean X bar is wanna wait and the sample standard deviation not the population standard deviation. The sample standard deviation is 10 and we're asked to find the 96% confidence interval whenever n equals 25. Now, since I only know s I only know the sample standard deviation, I need to use the tea interval instead of the Z interval. So you can do manually, I'm going to use the T I 84 because it doesn't really nice. So if you go to stat and then tests, it's going to be this eighth option down here, the T interval. So usually if you know sigma you're gonna use the Z interval, but this time I'm going to go to the eighth option, the tea interval, oops, Do it again, so stat tests and then eight. Okay, And then make sure you're covered over stats here, Um that's highlighted, and then you can start typing in your summary stats. So X. bar is 108. S. is 10 and then the sample size was 25 and then the sea levels .96. And whenever we calculate that this first line here that's going to be the uh confidence intervals. So one of 3.66 all the way up to 1 12 point 34. Okay so one of three 0.66 And 1 1234. Okay now we're gonna do the 96% confidence interval whenever we decrease the sample size to 10. So let's see what happens here. If we go back to stat And you can probably guess what's going to happen and then test and then we go down to this eighth option again and then everything stays the same except this time the n equals 10 And everything else stays the same. So we calculate and that gives us this confidence interval 100.42 all the way up. 215.58. Okay so let's write that down and compare 142 All the way up to 115.58. Okay. So what's the relationship between N. And the margin of air or the interval size? Well as you can see as in decreases the margin of air is going to increase. Making making the confidence interval wider. Okay. And this is super typical. So we've seen this before with the Z. Interval um The less you know or the smaller your sample size, the more variable that's gonna be especially the T. Distribution. You got to think of that. Um ah T distribution as in decreases that makes that t distribution more variable or more spread. Um So that's why that confidence intervals wider so that the lower bound is smaller and then the upper bound is a little bit higher. So what happens that's with the sample size. Now? What happens with the confidence interval this time we're back to the 25 is there in? But we need to find the 90% confidence interval. So we're gonna go back to stat Tests and then it's the 8th option again. And this time. So let's change this back to 25 And then the sea level we're going to change 2.9 for 90%. So we calculate what do we get? 1? 11.42. So let's write that down 104 0.58 All the way up to 111 42. So what's the relationship between the confidence level and the interval width? So if you compare this to part A it's a little bit narrower. Internet, so we go from one of 3 to 1 to force the lower bound is increased and then the upper bound 1 12 to 1 11 is decreased. So the interval the confidence interval is narrower than in part A. Okay, so this shows you that as confidence level decreases, the margin of error also decreases, making that that interval more narrow. Okay. And then the last part just asks um let's it says in the very very beginning that this population comes from approximately normal distribution. So could we conduct this test if the population were not normal? And the short answer would be no. If the population is not normal. Mhm. Then in needs to be at least 30 for Z or T. But in this case the T. Distribution. Okay, so since it's not normal, it's okay. If it is 25 or 10 or whatever, but if it's not normal, then that end needs to be at least 30 in order for us to use that central limit theorem and conduct this test.


Similar Solved Questions

5 answers
E27 Tzm9 m=ux T-Jax 7& =| 5 2Fz d-dx V = d + = 4 51/ + G.8 172 (9.8x _+=[gr3x
E27 Tzm9 m=ux T-Jax 7& =| 5 2Fz d-dx V = d + = 4 51/ + G.8 172 (9.8x _ += [gr3x...
5 answers
Prove that the following hypotheses: H1, H2,H3and H4 implies the propo- sition: T, whereH1: P ^ Q H2: P_Q^S, H3: R- S H4: T^ P _ R
Prove that the following hypotheses: H1, H2,H3and H4 implies the propo- sition: T, where H1: P ^ Q H2: P_Q^S, H3: R- S H4: T^ P _ R...
5 answers
F(z,y) = 5 + 76.V +38.2 + 240y
f(z,y) = 5 + 76.V +38.2 + 240y...
5 answers
1= Evaluate the following integrals: In(o) dr [A](d) K & cos(t w) dw sin(3 In €) dxtan -() dx[4]Te 4r dxIr 'In(z2 + 9) dx
1= Evaluate the following integrals: In(o) dr [A] (d) K & cos(t w) dw sin(3 In €) dx tan - () dx [4] Te 4r dx Ir 'In(z2 + 9) dx...
5 answers
0<0<27 Use numbers in exact form: Write {$- in the form :=r(cos0+isin0)'
0<0<27 Use numbers in exact form: Write {$- in the form :=r(cos0+isin0)'...
2 answers
Information follows about sums of squares and sample size for & linear model Use this information to fill in all values in an ANOVA table for regression Include your work and a shaded sketch 'showing the T.S. and p-value (10 pts ) SSModel 8.5 SSError 247.4 40Source_dfSSMSp-valucModelErrorTotalSketch
Information follows about sums of squares and sample size for & linear model Use this information to fill in all values in an ANOVA table for regression Include your work and a shaded sketch 'showing the T.S. and p-value (10 pts ) SSModel 8.5 SSError 247.4 40 Source_ df SS MS p-valuc Model...
5 answers
If 50.9 g of platinum, originally at 30.0*C absorbs 1,351 J of heat what is the final Celsius temperature of the platinum?
If 50.9 g of platinum, originally at 30.0*C absorbs 1,351 J of heat what is the final Celsius temperature of the platinum?...
5 answers
5 Find 841 Polnts) equivalent DETAILS resistance between Ci10 20.P.054. points and B in the drawiingcDock Addltionn1112.00 9Ila 1
5 Find 841 Polnts) equivalent DETAILS resistance between Ci10 20.P.054. points and B in the drawiing cDock Addltionn 1 1 1 2.00 9 Ila 1...
5 answers
Internal forces cannot change the total momentum of a system.True False
Internal forces cannot change the total momentum of a system. True False...
5 answers
Using Vtor and your value for RTOT; calculate the current I-Calculate Vt, the voltage drop across Ri:Vi10. Calculate Vz and Vs, the voltages across Rz and R;VzVz =11. Calculate Iz and I3_12. Calculate the sum:I2Write a formula relating I1 to Izand Iz :13. From 5 and above, calculate:Vi + Vz =Vi + Vz =What relationships become clear:14. Enter the results of your calculations in the table belowVa+Vz (v)(a)0 | v | v | (v)Calculated Results
Using Vtor and your value for RTOT; calculate the current I- Calculate Vt, the voltage drop across Ri: Vi 10. Calculate Vz and Vs, the voltages across Rz and R; Vz Vz = 11. Calculate Iz and I3_ 12. Calculate the sum: I2 Write a formula relating I1 to Izand Iz : 13. From 5 and above, calculate: Vi + ...
5 answers
Evaluate the double integral by first identifying it as the volume of a solid f 6 (8 _o)dA R={(I mio < o <8.0 < v < 3} The value of integral is (156
Evaluate the double integral by first identifying it as the volume of a solid f 6 (8 _o)dA R={(I mio < o <8.0 < v < 3} The value of integral is (156...
5 answers
A wire Of lengtn 9 1S cut into tWo pleces wnicn are then bent into tne shape ofa circle of radius r and a square of side $. Then the total area enclosed by the circle and square is the following function of and rIf we solve for $ in terms of r, we can reexpress this area as the following function of r alone:Thus we find that to obtain maximal area we should let r9 2TTo obtain minimal area we should let r(8 + 4n )
A wire Of lengtn 9 1S cut into tWo pleces wnicn are then bent into tne shape ofa circle of radius r and a square of side $. Then the total area enclosed by the circle and square is the following function of and r If we solve for $ in terms of r, we can reexpress this area as the following function o...
5 answers
Hercellis der d liketthexylethe companion cell shares functions with the sicve tube; including ATP production,nucleus and vacuole functionsIncorrectQuestion 20/1ptsWhich of these statements is false concerning phloem transport?incoina specics sorbilo and mannito can be transportedsugars like sucrose are transportcdpolymer trapping speciesuse long chain SuBAF potymers instead of sucroscrcducinc surars domninala[ronsport JUnarslacBook Pro30888
Hercellis der d liketthexyle the companion cell shares functions with the sicve tube; including ATP production,nucleus and vacuole functions Incorrect Question 2 0/1pts Which of these statements is false concerning phloem transport? incoina specics sorbilo and mannito can be transported sugars like ...
5 answers
The demand function for a certain product is p = 144 − x2 and the supply function is p = x2 + 2x + 132. Find the equilibrium point.(x, p) = Find the consumer's surplus there. (Round your answer to thenearest cent.)$
The demand function for a certain product is p = 144 − x2 and the supply function is p = x2 + 2x + 132. Find the equilibrium point. (x, p) = Find the consumer's surplus there. (Round your answer to the nearest cent.) $...
5 answers
Zhdlala Jzl a MF 3aa 2 slelal
zhdlala Jzl a MF 3aa 2 slelal...
5 answers
Pendulum released from point A becomes projectile at point B and lands at point C. Ih = 20 cm and y 0.5 m, what would be the velocity of the projectile when it hits the ground?"J" TuL "Uctsi' r]3eSe ),0n Wna^ Ucsq X< ~pjui "' (J(: : 7TpJMc (12 nAnug ~m{Oa. {"ebumsnh.
pendulum released from point A becomes projectile at point B and lands at point C. Ih = 20 cm and y 0.5 m, what would be the velocity of the projectile when it hits the ground? "J" TuL "Uctsi' r]3eSe ),0n Wna^ Ucsq X< ~pjui "' (J(: : 7TpJMc (12 nAnug ~m{Oa. {"eb...
5 answers
11. (10 pts) The graph of f (x) is given below. List the transformations, in order; to graph g(x) Then graph g(x) ~f(x 1) + 2 on the grid. Make sure to plot and label at least 2 points on the graph: Q11 4 Transformations: or right? (Hatoll shd L0 Jlbd_ qve( specify the axis uedkya( SLY | 26,3)SL4)0
11. (10 pts) The graph of f (x) is given below. List the transformations, in order; to graph g(x) Then graph g(x) ~f(x 1) + 2 on the grid. Make sure to plot and label at least 2 points on the graph: Q11 4 Transformations: or right? (Hatoll shd L0 Jlbd_ qve( specify the axis uedkya( SLY | 2 6,3) SL4)...

-- 0.019769--