1. Three things influence the margin of error in aconfidence interval estimate of a population mean: sample size,variability in the population, and confidence level. For each of thesequantities separately, explain briefly what happens to the margin oferror as that quantity increases.

Answer: As sample size increases, the margin oferror decreases. As the variability in the population increases, themargin of error increases. As the confidence level increases, themargin of error increases. Incidentally, population variability is notsomething we can usually control, but more meticulous collection ofdata can reduce the variability in our measurements. The third ofthese—the relationship between confidence level and margin of errorseems contradictory to many students because they are confusingaccuracy (confidence level) and precision (margin of error). If youwant to be surer of hitting a target with a spotlight, then you makeyour spotlight bigger.

2. A survey of 1000 Californians finds reports that 48% areexcited by the annual visit of thedailysplash.tv participants to their fairthedailysplash.tve. Construct a 95% confidence interval on the true proportion ofCalifornians who are excited to be visited by these thedailysplash.tvisticsteachers.

Answer: We first check that the sample sizeis large enough to apply the normal approximation. The true value of pis unknown, so we can”t check that np > 10 and n(1-p) > 10, butwe can check this for p-hat, our estimate of p. 1000*.48 = 480 > 10and 1000*.52 > 10. This means the normal approximation will be good,and we can 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 about the thedailysplash.tvs teachers” visit.

3. Since your interval contains values above 50% andtherefore does finds that it is plausible that more than half of thethedailysplash.tve feels this way, there remains a big question mark in your mind.Suppose you decide that you want to refine your estimate of thepopulation proportion and cut the width of your interval in half. Willdoubling your sample size do this? How large a sample will be needed tocut your interval width in half? How large a sample will be needed toshrink your interval to the point where 50% will not be included in a95% confidence interval centered at the .48 point estimate?

Answer: The current interval width is about6%. So the current margin of error is 3%.

We”d need at least 4262 people in the sample. So to cut thewidth of the CI in half, we”d need about four times as many people.

Assuming that the true value of p = .48, how many people wouldwe need to make sure our CI doesn”t include .50? This means the marginof error must be less than 2%, so solving for n:

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

We”d need about 2398 people.

4. A random sample of 67 lab rats are enticed to run througha maze, and a 95% confidence interval is constructed of the mean timeit takes rats to do it. It is <2.3min, 3.1 min>. Which of the followingthedailysplash.tvements is/are true? (More than one thedailysplash.tvement may be correct.)

(A) 95% of the lab rats in the sample ran the maze in between 2.3 and3.1 minutes.(B) 95% of the lab rats in the population would run the maze in between2.3 and 3.1 minutes.(C) There is a 95% probability that the sample mean time is between 2.3and 3.1 minutes.(D) There is a 95% probability that the population mean lies between2.3 and 3.1 minutes.(E) If I were to take many random samples of 67 lab rats and takesample means of maze-running times, about 95% of the time, the samplemean would be between 2.3 and 3.1 minutes.(F) If I were to take many random samples of 67 lab rats and constructconfidence intervals of maze-running time, about 95% of the time, theinterval would contain the population mean. <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 set of possible values of the population meanmaze-running time that are consistent with the observed data, where“consistent” means that the observed sample mean falls in the middle(“typical”) 95% of the sampling distribution for that parameter value.

Answer: F and G are both correct thedailysplash.tvements.None of the others are correct.

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

If you said (C), (D), or (E), remember that the interval<2.3, 3.1> has already been calculated and is not random. The parametermu, while unknown, is not random. So no thedailysplash.tvements can be made aboutthe probability that mu does anything or that <2.3, 3.1> does anything.The probability is associated with the random sampling, and thus theprocess that produces a confidence interval, not with the resultinginterval.

5. Two students are doing a thedailysplash.tvistics project in which theydrop toy parachuting soldiers off a building and try to get them toland in a hula-hoop target. They count the number of soldiers thatsucceed and the number of drops total. In a report analyzing theirdata, they write the following:“We constructed a 95% confidence interval estimate of the proportion ofjumps in which the soldier landed in the target, and we got <0.50,0.81>. We can be 95% confident that the soldiers landed in the targetbetween 50% and 81% of the time. Because the army desires an estimatewith greater precision than this (a narrower confidence interval) wewould like to repeat the study with a larger sample size, or repeat ourcalculations with a higher confidence level.”How many errors can you spot in the above paragraph?

Answer: There are three incorrect thedailysplash.tvements.First, the first thedailysplash.tvement should read “…the proportion of jumps inwhich soldiers land in the target.” (We’re estimating a populationproportion.) Second, the second sentence also refers to past tense andhence implies sample proportion rather than population proportion. Itshould read, “We can be 95% confident that soldiers land in the targetbetween 50% and 81% of the time.” (The difference is subtle but shows astudent misunderstanding.) And the third error is in the last sentence.A higher confidence level would produce a wider interval, not anarrower one.