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1 Sectio 10 Aswer Key: % 60% 70% 80% 90% 95% 96% 98% 99% 99.5% 99.8% 99.9% 1) A simple radom sample of New Yorkers fids that 87 are left-haded. (a) Fid the 95% cofidece iterval for the proportio of New Yorkers who are left-haded. p = 87 = p ± z = ± ± = (0.0695,0.1045) We are 95% cofidet that the proportio of New Yorkers who are left-haded is with a margi of error of (b) Fid the 99% cofidece iterval: p ± z = ± ± = ( , ) We are 99% cofidet that the proportio of New Yorkers who are left-haded is with a margi of error of (c) We ca be 99.9% cofidet that the proportio of New Yorkers who are left-haded is betwee what two umbers? p ± z = ± ± = (0.0577,0.1163) We are 99.9% cofidet that the proportio of New Yorkers who are left-haded is with a margi of error of (d) Either our group of is amog the 10% most uusual samples, or the proportio of New Yorkers who are left-haded is betwee what two umbers?

2 p ± z = ± ± = (0.072,0.102) We are 90% cofidet that the proportio of New Yorkers who are left-haded is with a margi of error of ) We wish to kow the probability that a suspect coi will lad heads. We flip the coi times ad 190 times it lads heads. (a) Fid the 80% cofidece iterval for the probability that the coi will lad heads. p = 190 = p ± z = ± ± = (0.443,0.507) (b) We ca be 95% cofidet that the probability of the coi ladig heads is betwee what two umbers? p ± z = ± ± = (0.426,0.475) (c) Either our sample of flips was amog the 1% most uusual, or that the probability of the coi ladig heads is betwee what two umbers? This is a 99% cofidece iterval: p ± z = ± ± = (0.411,0.539) 3) Pickig 250 orders radomly from a mail-orderig compay, we fid that 210 arrived o time. Let p be the proportio of all orders that are o time. (a) Fid the 98% cofidece iterval for p. (b) p = = 0.84 p ± z = 0.84 ± ± = (0.786,0.894) We ca be 99% cofidet that p is betwee what two umbers?

3 p ± z = 0.84 ± ± = (0.780,0.900) (c) Either our sample of 250 orders was amog the 5% most uusual, or p is betwee what two umbers? This is a 95% cofidece iterval: p ± z = 0.84 ± ± = (0.795,0.885) 4) I our effort to fid out what percetage of all statistics are meaigless, we do a well-fuded study ad lear that of 420 examied statistics, 386 of them were meaigless. Fid a 99.5% cofidece iterval for the true proportio of all statistics that are meaigless. p = = ( ) p ± z = ± ± = (0.882,0.956) 5) The U.S.R.S. (Uio of Starfleet Red Shirts) wats to kow the probability of a Red Shirt dyig whe he beams dow to a plaet with Captai Kirk. Fid a 99.8% cofidece iterval for this probability, after learig that 35 of the last 93 Red Shirts who beamed dow with Kirk met a ufortuate ed. p = = ( ) p ± z = ± ± = (0.221,0.531) 6) A radom sample of 1021 adults foud that 38% said they believe i ghosts. Fid a 90% cofidece iterval for the percetage of all adults who believe i ghosts. Fid a 99% cofidece iterval. p = (1 0.38) p ± z = 0.38 ± ± = (0.3545,0.4055) 0.38(1 0.38) p ± z = 0.38 ± ± = (0.341,0.419)

4 7) A isurace compay checks police records o 582 accidets selected at radom ad otes that teeagers were at the wheel i 91 of them. Fid a 95% cofidece iterval for the true proportio of all auto accidets that ivolve teeage drivers. p = = ( ) p ± z = ± ± = (0.126,0.186) 8) A compay wats to test the respose to a ew flier, ad they sed it to people radomly selected from their mailig list of over 200,000. They get orders from 123 of the recipiets. Create a 90% cofidece iterval for the percetage of people the compay cotacts who will sed i a order. The full mailig wo t be cost effective uless it produces at least a 5% retur. Should they do it or ot? p = 123 = ( ) p ± z = ± ± = (0.106, 0.140) Oe of three thigs is the case: either our sample is amog the 90% most typical (i which case p, the proportio of the etire mailig list which will sed i a order, is betwee ad 0.140), or it's amog the 10% outlyig tails. If it's amog the 5% lowest samples (a uusually low sample proportio uderestimatig p) the p is more tha It is of course fie if more tha 14% will sed i orders. If it's amog the 5% highest samples (a uusually high sample proportio overestimatig p) the p is less tha But eve so, it could still be much higher tha the 0.05 which they require. So we ca be very cofidet that p is more tha the 0.05 which makes it worth their while. 9) It s believed that as may as 25% of adults over 50 ever graduated from high school. We wish to see if this percetage is the same amog the 25 to 30 age group. (a) How may of this youger age group must we survey i order to estimate the proportio of o-grads to withi 6% with 90% cofidece? For 90% cofidece, our z is 1.645, our margi of error is 0.06, ad we require a p to use. Although the 0.25 refers to a differet populatio, it at least is a wild guess tha we ca use, sice we do't kow our p, the proportio of youg adults who ever graduated, or of course do we kow p, the outcome of the sample whose size we're still tryig to determie: = ( z m )2 p(1 p) = ( ) 0.25(1 0.25) = 140.9, so = 141. Alterately, if we wat to use p=0.5 (coverig the worse-case sceario which overestimates) we get = 188. Which is better? I fact, these two types of people are probably ot very similar regardig graduatio rate, so our wild guess is pretty wild. But because it's more cetral (closer to 0.5) tha the truth (likely less tha 0.25 of youg adults failed to graduate; i.e., p is more extreme), really the p=0.25 will overestimate as well, just ot as badly as the p=0.5.

5 (b) Suppose we wat to cut the margi of error dow to 4%. How may? = ( z 2 m ) p(1 p) = ( ) 0.25(1 0.25) = 317.1, so = 318. (rememberig that we have to roud up to the smallest acceptable sample size). (c) Dow to 3% how may? = ( z m )2 p(1 p) = ( ) (1 0.25) = 563.8, so = 564.

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