4-8 - Chapter 10 — Estimating with Confidence l Example...

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Unformatted text preview: Chapter 10 — Estimating with Confidence l Example 2: Voting g In a random sample of 300 students at a local university that has 20,000 students, 20% 1‘ of the sampled students said they were not going to vote in the fall elections. Find a 95% confidence interval for the proportion p of college students at the university who are not going to vote in the fall. @ I5\\\l>7_o.mf Yeé \/ (sfmfaler 6E) Q) CL—rg {H8 : 3002352. : (002M) yas/ (CI halal) A . F) : (5. Z '- ZgQ ‘1 (weM/e alone 'lE’r'lrfiqlrquiy) siege) : lease. 1 w ’ . “E5” 'fi§:*“echozat 95570 CI: )8 1 i*,§E(1B>: a2 4' * 1 Example 3: Bar ,revisited * > ' - ' Recall from Exampl 0 in Chapter 9 that in a n om sample of 250 residents of a city with at least 10,000 people, 114 people in the sample said they favor raising the bar entrance age from 19 to 21. Find a 80% confidence interval (CI) for that city’s population proportion p of people who favor raising the bar age. (D and @ beefs (Jame {‘n C kqu‘er Ci ,, fl: “pl/.250 : V O<Z Ogl 7:7 M/ZZO/l * . 13, = ZvX/‘Z’ = 2a,, : LZ87. {313(‘9 : v,w5e—Cz-—,Lle@/Yzeo ’96 aegis 800),) CI; ,3: 2551549) : O‘Elf'aéierBZ'aoglb 'v lEbéz 0.4%,! Example 4: DrinkingAge (6‘ q -) 1) In a random sample of 350 students at a local university (the university has 0 approximately 15,000 students), 60% said they were in favor of lowering the drinking age from 21 to 18. Find. a 90% CI for the proportion p of college students at the uversity who favor lowering the drinking age from 21 to 18. (D Is N > law/L Yes. \/ (no Correction @6165") /@ CLTZ: asp“ : 3502M, : z/oz/o ;fi_‘(l§fg\) 2 3.500%: I‘lozl‘o /\ [3:062 0(r/‘“O,Cl:©.l 22*: Z07; 2 E00 : Mott?) Eb 649$): {Mg/Wt : \m/l/E 20.62422 4 (-10%) CI: fit 2*, 52943185 ‘ Oéilfid§%MfléZrfl@&55éq)offi3h) l amoebicinitiioioiioiiaiiaafiattaetiaéaattafifi Chapter 10 — Estimating with Confidence 10.2 Confidence Intervals for a Population Mean u when 0' is Unknown Assume that the random observations X1, X2, ..., Xn are either repeated, independent measurements or a simple random sample from a many times larger reai population. Also, assume the sampling distribution of the sample mean )7 is close to normal. If (5 is known, then for some positive constant 2*, the confidence interval for p is A7 i 2* i. l y; However, it is often the case that o is unknown, so we will focus on this situation. When 6 is unknown, then for some positive constant t*, the confidence interval for p is L _ . 5 ' ’ X i t —— , " \//7 where t* refers to the t-curve. For X1, X2, ..., Xn independent observations from a normal population with mean u and standard deviation 0', the one-sample standardized t-statistic, X—u S/n' has a so—called t—distribution with n — 1 degrees of freedom (df). The degrees of freedom are used to identify the particular t-distribution. - 7': Each t-curve shares the following properties of the standard normal curve: 0 a The curve is bell—shaped and symmetric around 0. o The curve extends indefinitely in both directions, always above the horizontal axis, approaching but never touching the axis. 0 The total area under the curve is 1. Example 5: Calculating t* Use Table F to find the critical value t* that corresponds to both the confidence intervals and the values of n that are provided. b . 4/1 ,4: 0,015 a) 95% CI for-n = 10. 150615 KEY: 0‘ :7 [fiaozg o = - W; 0‘ \ b) 90 /0 CI for n 15. Cf—P: '4 Etc-5.050%] «:0,2‘ c)80°/oCIforn_='20. _ g ' drzn =7 («326 it... 0601 106' Chapter 10 —- Estimating with Confidence Even if the distribution is skewed, if the sample size is large enough, we can still use the t-curve to construct confidence intervals. Example 6: Average Weight V 81 Suppose the average weight of a random sample of E men between the ages of 18 and 22 was )7 = 170.3 lb with a standard deviation of s = 36.4 lb. Construct a 90% CI for the average weight of all men between the ages of 18 and 22. x__. X : 170.3 lb J74: 8l~l : 8Q 75* : rem [twgaofl m/(eelqcemwt‘b (M > 11> :40 COCWGde/Léc‘tmf _ 5 . b ‘ . 36.- sax): 77:: §% 41;: mgr/5114,54 «gt—q : l 3,5 '77 ,0 Example 7: Working Part-Time5, ( 6’ 7 I 3) In a survey of random sample offi undergraduate students at a large university, it was ' found that the mean amount of time that the students spend working at a part-time job (during the school year) was )7 = 22.4 hours per Week with a sample standard deviation of s = 2.1 hours. Find a 95% CI for the mean amount of time that all undergraduates spend working at part-time jobs. ' (’3 1.6 N > ZO~fL Z \{66 C r’lo Corr¢¢+l\on \C'QOLal‘) 01.18 : 551' l --— £50 I) Y». 2234 wig/wk _ . _, g_ 2.1 75*; fd.ozé(50>- 200% ‘3 .‘gECX> 2 V7; ’ _ g _ ._ 955% CI: X it)‘ FF: :22.HIZ.C@GZ. A 53 Example 8: Average Height 2'? at > life /LC: Suppose the average height of a random sample of wig/Pmen between t e ages 0 and 22 was Y = 62.2 inches with a standard deviatio of s = 1.4 inches. Construct an \N ® 80% CI for the average height of all women between the ages of 18 and 22. g (D Ié N> 'w‘fLZ Yes (no corrécti‘ol/I‘der> e 4%.;8/wzso ‘ 72692.2m Q 94 , . 2 fl __ ll‘i in a t : alwayrzqi, 6Eéx>\m-@ l l ...
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4-8 - Chapter 10 — Estimating with Confidence l Example...

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