Confidence Intervals

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Solutionsto Practice Problems

1. Three things affect the margin of error in aconfidence interval estimate of a population mean: sample dimension,variability in the populace, and also confidence level. For each of thesequantities independently, describe briefly what happens to the margin oferror as that amount boosts.

Answer: As sample dimension boosts, the margin oferror decreases. As the varicapability in the populace rises, themargin of error rises. As the confidence level boosts, themargin of error boosts. Incidentally, population varicapability is notsomething we deserve to generally regulate, yet even more meticulous collection ofdata deserve to alleviate the varicapacity in our measurements. The third ofthese—the connection in between confidence level and margin of errorseems inconsistent to many type of students bereason they are confusingaccuracy (confidence level) and also precision (margin of error). If youwant to be surer of hitting a target through a spotlight, then you makeyour spotlight bigger.

2. A survey of 1000 Californians finds reports that 48% areexcited by the annual visit of participants to their fairnlinux.orge. Construct a 95% confidence interval on the true propercentage ofCalifornians who are excited to be went to by these nlinux.orgisticsteachers.

Answer: We first inspect that the sample sizeis huge enough to apply the normal approximation. The true worth of pis unwell-known, so we can not check that np > 10 and n(1-p) > 10, butwe have the right to check this for p-hat, our estimate of p. 1000*.48 = 480 > 10and also 1000*.52 > 10. This means the normal approximation will be great,and we have the right to apply them to calculate a confidence interval for p.

.48 +/- 1.96*sqrt(.48*.52/1000)

.48 +/- .03096552 (that mysterious 3% margin of error!)

(.45, .51) is a 95% CI for the true proportion of allCalifornians who are excited around the nlinux.orgs teachers" visit.

3. Due to the fact that your interval contains values over 50% andtherefore does finds that it is plausible that even more than fifty percent of thenlinux.orge feels this method, there continues to be a big question mark in your mind.Suppose you decide that you desire to refine your estimate of thepopulation propercentage and also cut the width of your interval in fifty percent. Willdoubling your sample dimension carry out this? How huge a sample will be necessary toreduced your interval width in half? How big a sample will certainly be necessary toshrink your interval to the allude where 50% will not be consisted of in a95% confidence interval centered at the .48 allude estimate?

Answer: The current interval width is about6%. So the existing margin of error is 3%. We want margin of error =1.5% or

1.96*sqrt(.48*.52/n) = .015

Solve for n: n = (1.96/.015)^2 * .48*.52 = 4261.6

We"d need at leastern 4262 human being in the sample. So to cut thewidth of the CI in fifty percent, we"d require about 4 times as many type of world.

Assuming that the true worth of p = .48, just how many kind of world wouldwe must make certain our CI does not encompass .50? This implies the marginof error must be much less than 2%, so fixing for n:

n = (1.96/.02)^2 *.48*.52 = 2397.1

We"d require around 2398 world.

4. A random sample of 67 lab rats are enticed to run througha maze, and a 95% confidence interval is built of the intend timeit takes rats to carry out it. It is <2.3min, 3.1 min>. Which of the followingnlinux.orgements is/are true? (More than one nlinux.orgement may be correct.)

(A) 95% of the lab rats in the sample ran the maze in in between 2.3 and3.1 minutes.(B) 95% of the lab rats in the populace would run the maze in between2.3 and also 3.1 minutes.(C) There is a 95% probcapacity that the sample expect time is between 2.3and also 3.1 minutes.(D) Tbelow is a 95% probability that the populace mean lies between2.3 and also 3.1 minutes.(E) If I were to take many random samples of 67 lab rats and also takesample implies of maze-running times, about 95% of the moment, the samplemean would certainly be in between 2.3 and 3.1 minutes.(F) If I were to take many type of random samples of 67 lab rats and constructconfidence intervals of maze-running time, around 95% of the time, theinterval would certainly contain the populace suppose. <2.3, 3.1> is the one suchpossible interval that I computed from the random sample I actuallyobserved.(G) <2.3, 3.1> is the collection of possible values of the population meanmaze-running time that are continual via the oboffered data, where“consistent” indicates that the oboffered sample suppose drops in the middle(“typical”) 95% of the sampling circulation for that parameter value.

Answer: F and also G are both correct nlinux.orgements.Namong the others are correct.

If you said (A) or (B), remember that we are estimating aintend.

If you said (C), (D), or (E), remember that the interval<2.3, 3.1> has actually currently been calculated and also is not random. The parametermu, while unknown, is not random. So no nlinux.orgements deserve to be made aboutthe probcapacity that mu does anypoint or that <2.3, 3.1> does anything.The probability is associated through the random sampling, and for this reason theprocedure that produces a confidence interval, not with the resultinginterval.

5. Two students are doing a nlinux.orgistics task in which theydrop toy parachuting soldiers off a structure and also try to get them toland in a hula-hoop tarobtain. They count the number of soldiers thatsucceed and also the variety of drops full. In a report analyzing theirinformation, they compose the following:“We constructed a 95% confidence interval estimate of the proportion ofjumps in which the soldier landed in the target, and we gained <0.50,0.81>. We deserve to be 95% confident that the soldiers landed in the targetin between 50% and also 81% of the time. Since the army desires an estimatevia higher precision than this (a narrower confidence interval) wewould certainly choose to repeat the study with a bigger sample size, or repeat ourcalculations with a higher confidence level.”How many errors have the right to you spot in the over paragraph?

Answer: Tright here are three incorrect nlinux.orgements.First, the first nlinux.orgement must read “…the propercent of jumps inwhich soldiers land also in the targain.” (We’re estimating a populationpropercentage.) 2nd, the second sentence additionally refers to previous tense andfor this reason indicates sample propercent fairly than populace proportion.

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Ithave to read, “We can be 95% confident that soldiers land also in the targetin between 50% and also 81% of the time.” (The difference is subtle however shows astudent misknowledge.) And the 3rd error is in the last sentence.A better confidence level would produce a broader interval, not anarrower one.