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1 Section 7.2 Check Your Understanding, age 445: 1. The mean of the samling distribution of ˆ is equal to the oulation roortion. In this case µ ˆ = = The standard deviation of the samling distribution of ˆ is ) 0.75( 0.25) ˆ = = = There are more than 10(1000) = 10,000 young adult n 1000 Internet users, so the 10% condition has been met. n = = 750and 3. The samling distribution of ˆ is aroximately Normal. Both n ) = 1000( 0.25) = 250are at least If the samle size were 9000 instead of 1000, the samling distribution would still be aroximately Normal with mean However, the standard deviation of the samling distribution would be smaller by a factor of 3. In this case, ) 0.75( 0.25) ˆ = = = n 9000 Exercises, age 447: 7.27 (a) We would not be surrised to find 8 (32%) orange candies. From the grah in Figure 7.11 there were a fair number of samles in which there were 8 or fewer orange candies. On the other hand, there were only a coule of samles in which there were 5 (20%) or fewer orange candies. So, if we got only 5 orange candies, we should be surrised. (b) It is more surrising to get 32% orange candies in a samle of 50 than it is in a samle of 25. Comaring the grahs in figures 7.11 and 7.12, there were a fair number of simulations in 7.11 (samle size 25) with 32% or less, but very few in 7.12 (samle size 50) with 32% or less (a) We would be surrised to find 32% orange candies in this case. Very few of the samles of size 25 had 32% or more orange candies. However, we would not be surrised to find 20% orange candies. 20% is very near the center of the distribution and not unusual at all. (b) We would be surrised to find 32% orange candies in either case since neither simulation had many samles with 32% or more orange candies. However, it would be even more surrising when the samle size is 50 because there were fewer samles of size 50 that had at least 32% orange candies than samles of size (a) The mean of the samling distribution is equal to the oulation roortion, so µ ˆ = = (b) The standard deviation of the samling distribution is (1 ) 0.45(0.55) ˆ = = = The 10% condition is met because it is very likely n 25 true that there are more than 10(25) = 250 candies in the large machine. n = = 11.25and (c) The samling distribution is aroximately Normal because n ) = 25( 0.55) = 13.75are both at least 10. Chater 7: Samling Distributions

2 (d) If the samle size were 100 rather than 25, the samling distribution would still be aroximately Normal with a mean of µ = However, the standard deviation decreases to ) 0.45( 0.55) ˆ ˆ = = = n (a) The mean of the samling distribution is equal to the oulation roortion, so µ ˆ = = (b) The standard deviation of the samling distribution is ) 0.15( 0.85) ˆ = = = In this case, the 10% condition is met because it is n 25 very likely true that there are more than 10(25) = 250 candies in the large machine. n = = 3.75is less (c) The samling distribution is not aroximately Normal because than 10. Note that n( 1 ) 25( 0.85) = = is at least 10, but the Large Counts conditions requires that both numbers must be at least 10. (d) If the samle size were 225 rather than 25, the samling distribution would now be n 1 = = are aroximately Normal because n = = and both at least 10. The mean is still µ ˆ = 0.15, but the standard deviation decreases to ) 0.15( 0.85) ˆ = = = n (a) No. The 10% condition is not met here because more than 10% of the oulation (10/76 = 13%) was selected. (b) No. The Large Counts condition is also not met because the samle size was only n = 10. n 1 will be at least 10. Neither n nor 7.32 (a) Yes. The 10% condition is met here because less than 10% of the oulation (7/100 = 7%) was selected. (b) The Large Counts condition is not met here because the samle size was only n = 7. Neither n 1 will be at least 10. n nor 7.33 The Large Counts condition is not met because n = 15( 0.3) = 4.5 < The 10% condition is not met because more than 10% of the oulation (50/316 = 15.8%) was selected (a) The mean of the samling distribution of ˆ is equal to the oulation roortion. In this case, µ ˆ = = (b) The standard deviation of the samling distribution of ˆ is (1 ) 0.7(0.3) ˆ = = = The 10% condition is met because the samle of size n is less than 10% of the oulation of all U.S. adults. (c) The samling distribution of ˆ is aroximately Normal because n = 1012( 0.70) = and n ) = 1012( 0.30) = 303.6are both at least 10. The Practice of Statistics for 5/e

3 (d) Ste 1: State the distribution and values of interest. ˆ has an aroximately Normal distribution with mean 0.70 and standard deviation We want to find P( ˆ 0.67). Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is P(Z 2.08) = Using technology: normalcdf(lower: 1000, uer: 0.67, µ : 0.70, : ) = Ste 3: Answer the question. There is a robability of obtaining a samle in which 67% or fewer say they drink the milk. Because this is a small robability, there is convincing evidence against the claim it isn t lausible to get a samle roortion this small by chance alone (a) The mean of the samling distribution of ˆ is equal to the oulation roortion. In this case, µ ˆ = = 0.4. (b) The standard deviation of the samling distribution of ˆ is ˆ = 0.4( 0.6) = The % condition is met because the samle of size 1785 is less than 10% of the oulation of adults. n = = 714 and (c) The samling distribution of ˆ is aroximately Normal because n ) = 1785( 0.6) = 1071 are both at least 10. (d) Ste 1: State the distribution and values of interest. ˆ has an aroximately Normal distribution with mean 0.40 and standard deviation We want to find P( ˆ 0.44). Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is P(Z 3.45) = = Using technology: normalcdf(lower: 0.44, uer: 1000, µ : 0.40, : ) = Ste 3: Answer the question. There is a robability of obtaining a samle in which 44% or more say they attended church last week. Because this is a small robability, there is convincing evidence against the claim it isn t lausible to get a samle roortion this large by chance alone Because the standard deviation is found by dividing by n, using 4n for the samle size halves the standard deviation ( 4n 2 n) = ; we would need to samle = 4048adults Because the standard deviation is found by dividing by n, using 9n for the samle size reduces the standard deviation of the samling distribution to one-third of the revious value ( 9 n 3 n) = ; we would need to samle = 16,065adults. Chater 7: Samling Distributions

4 7.39 Ste 1: State the distribution and values of interest. We have an SRS of size 267 drawn from a oulation in which the roortion = 0.70of college women have been on a diet within the ast 12 months. Thus, µ ˆ = Because 267 is less than 10% of the oulation of college 0.7( 0.3) ˆ = = Because women, n = = and 267 n 1 = = 80.1are both at least 10, the samling distribution of ˆ can be aroximated by a Normal distribution. We want to find ( ˆ 0.75) P using the N(0.70, ) distribution. Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is PZ 1.79 = = Using technology: normalcdf (lower: 0.75, uer: 1000, µ : 0.7, : ) = Ste 3: Answer the question. There is a robability that 75% or more of the women in the samle have been on a diet within the last 12 months Ste 1: State the distribution and values of interest. We have an SRS of size 500 drawn from a oulation in which the roortion = 0.14 of registered motorcycles are Harley- Davidsons. Thus, µ ˆ = Because 500 is less than 10% of the oulation of registered 0.14( 0.86) ˆ = = Because motorcyles, n = = 70 and 500 n 1 = = 430are both at least 10, the samling distribution of ˆ can be aroximated by a Normal distribution. We want to find ( ˆ 0.20) P using the N(0.14, ) distribution. Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is P( Z 3.87) 0. Using technology: normalcdf (lower: 0.20, uer: 1000, µ : 0.14, : ) = Ste 3: Answer the question. There is a robability that 20% or more of the motorcycles in the samle are Harleys (a) Ste 1: State the distribution and values of interest. We have an SRS of size 100 drawn from a oulation in which the roortion = 0.90 of orders are shied within three working days. Thus, µ ˆ = Because 100 is less than 10% of the oulation of orders 0.90( 0.10) ˆ = = Because (100/5000 = 2%), n = = 90 and 100 n 1 = = 10are both at least 10, the samling distribution of ˆ can be aroximated by a Normal distribution. We want to find ( ˆ 0.86) P using the N(0.90, 0.03) distribution. Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is P( Z 1.33) = Using 0.03 technology: normalcdf (lower: 1000, uer: 0.86, µ : 0.90, : 0.03) = Ste 3: Answer the question. There is a robability that 86% or fewer of orders in an SRS of 100 were shied within 3 working days. The Practice of Statistics for 5/e

5 (b) It isn t unusual to get a samle roortion of 0.86 or smaller when selecting an SRS of 100 from a oulation in which = Thus, it is lausible that the 90% claim is correct and that the lower than exected ercentage is due to chance alone (a) Ste 1: State the distribution and values of interest. We have an SRS of size 100 drawn from a oulation in which the roortion = 0.67 of college students suort a crackdown on underage drinking. Thus, µ ˆ = Because 100 is less than 10% of the oulation of students at a large college, ˆ = = Because 100 n 1 = = 33are both at least 10, the samling distribution n = 100( 0.67) = 67 and of ˆ can be aroximated by a Normal distribution. We want to find ( ˆ 0.62) P using the N(0.67, ) distribution. Ste 2: Perform calculations. Show your work. The standardized score for the boundary value is z = = The desired robability is P Z 1.06 = Using technology: normalcdf (lower: 1000, uer: 0.62, µ : 0.67, : ) = Ste 3: Answer the question. There is a robability that 62% or fewer students in an SRS of 100 suort a crackdown on underage drinking. (b) It isn t unusual to get a samle roortion of 0.62 or smaller when selecting an SRS of 100 from a oulation where = Thus, it is lausible that the 67% claim is correct and that the lower than exected ercentage is due to chance alone a b b b The Venn diagram is shown below. 62% neither download nor share music files (a) Assign numbers to the animals (01 to the desert tortoise, 02 to the Olive Ridley sea turtle,, 14 to the San Francisco garter snake). Starting at line 111 in Table D, read airs of numbers until you get three different numbers between 01 and 14. These numbers reresent the animals chosen. (b) Using line 111 in Table D, the animals chosen are 12, 04, and 11, which reresent the blunt-nosed leoard lizard, the flat-tailed horned lizard and the Coachella Valley fringe-toed lizard. Chater 7: Samling Distributions