650.Homework

Unit 1 1.3 A. Download time is considered a continuous numerical variable / quantitative because it can take any real value within a specific range. In other words, there is no restriction on the possible values the download time can have, and it can be measured with great precision, even beyond the decimal point. B. The download time is considered a ratio-scaled variable because it possesses all the properties of a numerical variable while also having a meaningful and absolute zero point. A ratio-scaled variable allows for the meaningful interpretation of ratios between measurements. 1.6 a. Categorical, nominal scale b. Numerical, continuous, ratio scale c. Categorical, nominal scale d. Numerical, discrete, ratio scale e. Categorical, nominal scale 1.7 a. Numerical, ratio scale, continuous b. Categorical, nominal scale c. Categorical, nominal scale d. Numerical, ratio scale, discrete 1.23 a. Given the availability of a comprehensive list comprising full-time students, several sampling approaches could be employed. For instance, a straightforward random selection of 200 students could be undertaken. In the event that student contentment regarding campus life, which might undergo random variations across the student body, is of interest, an alternative method involves systematically selecting every twentieth student from the population frame. Should the investigation extend to potential variations in student contentment based on gender and academic standing, a more intricate approach may be warranted. This could entail employing a stratified sampling technique utilizing eight distinct strata encompassing female freshmen through female seniors, as well as male freshmen through male seniors. In situations where the hypothesis suggests that fluctuations in student satisfaction with campus life are comparable both within and between certain clusters, a cluster sampling strategy becomes viable. This would involve selecting specific groups for study rather than individual students. b. Selecting a simple random sample stands as one of the most straightforward methods. The pool of potential participants comprises a list of 4,000 student names held within the registrar's records. c. Choosing a systematic sample from the registrar's records is more convenient for manual selection compared to a simple random sample. In this approach, an initial individual is

randomly chosen, and subsequently, every twentieth person is included in the sample. Additionally, utilizing a systematic sample would offer the advantage of aligning the alphabetical distribution of the sampled students' names more closely with the alphabetical distribution of student names within the campus population. d. In the presence of accessible gender and class-specific lists, it is advisable to opt for a stratified sample. Given the potential divergence in student contentment based on gender and academic standing, employing a stratified sampling approach accomplishes two key objectives. Firstly, it guarantees the inclusion of all defined strata within the sample. Secondly, it enhances the representativeness of the sample, leading to more accurate estimations of population parameters with improved precision. e. In the scenario where all 4,000 full-time students are accommodated within 10 on-campus residence halls that integrate students based on both gender and class, it is recommended to utilize a cluster sampling method. In this approach, a cluster can be defined as a complete residence hall, and a sample can be obtained by selecting one residence hall at random and sampling the students within it. Given that each dormitory houses 400 students, a systematic sample of 200 students can be drawn from the selected cluster of 400 students. Alternatively, another cluster definition could involve selecting an entire floor within one of the 10 dormitories. Assuming each dormitory has four floors with 100 students on each floor, a viable approach would be to randomly choose two floors, resulting in the desired 200 student sample. Opting for the entire dormitory as a cluster may facilitate the distribution and collection of the survey, yet if there exists a variable beyond gender or class that varies among dormitories, selecting by floor might yield a more comprehensive and representative sample. 1.31 a. Possible coverage error: Sampling was limited to employees within a particular division of the company. b. Possible nonresponse error: There was no attempt to get nonrespondents to answer/ finish c. Possible sampling error: There was ambiguous wording in questions asked on the questionnaire. d. The statistical measures derived from the sample will differ from the population's underlying parameters of interest. Unit 2 3.4

b. 38/296 = 0.128 c. (150 + 58 - 38)/296 = 170/296 = 0.574 d. The likelihood of "being a female SMB owner or having concerns about limited access to money impacting the business" encompasses the likelihood of "being a female SMB owner" and the likelihood of "having concerns about limited access to money impacting the business," minus the combined likelihood of "being a female SMB owner and having concerns about limited access to money impacting the business." 4.14 a. 1,106/1,261 = 0.877 b. 911/1,261 = 0.7224 c. 1,001 + 1,106 - 911/1,261 = 1,196/1,261 = 0.949 d. 1,261/1,261 = 1.00 4.16 a. P(A|B)= 10/30 = ⅓ = 0.33 b. P(A|B’)= 20/60 = ⅓ = 0.33 c. P(A’|B’)= 40/60 = ⅔ = 0.67 d. Because P(A|B) = P(A)= ⅓ this means that events A and B are statistically independent. 4.22 a. 1,266/1,947 = 0.6502 b. 1,625/ 3,779 = 0.43 c. Since P(increased use of LinkedIn)= 0.5049 and P(increased use of LinkedIn | B2B marketer)= 0.6502, increased use of LinkedIn and business focus are not independent because they do not equal each other. 4.26 a. 0.025/0.6 = 0.0417 b. 0.015/0.4= 0.0375 c. Since 0.025/0.6 = 0.0417, and is not equal to 0.4, the two events are not independent. 5.2 a. The mean number of accidents per day is 2. b. The standard deviation is 1.18321596 c. P(X ≥2) = 0.45 + 0.15 + 0.05 + 0.05 = 0.7 d. The cumulative prize expense amounts to $25,000 + $100 + 31,476 * S5, totaling $182,480. If we consider the production cost of the flyers to be insignificant, the expenditure to reach a solitary customer comes to $182,480/31,478, resulting in $5.80. The success of the promotion hinges on the attendance count of customers at the showroom. 5.5 a. μ=E(X) = 2.9 b. σ = 1.77 c. P(X ﹤ 2) = 0.07 + 0.155 = 0.225

5.7 a. E(X) = $71 E(Y) = $97 b. σx = 61.88 σy = 84.27 c. Stock Y gives the investor a higher expected return than stock X, but also has a higher standard deviation. Risk-averse investors would invest in stock X, whereas risk takers would invest in stock Y. Unit 4 6.1 a. P(Z < 1.57) = 0.9418 b. P(Z> 1.84) = 1 - 0.9671 = 0.0329 c. P1.57<Z<1.84) = 0.9671-0.9418 = 0.0253 d. P(Z < 1.57) + P(Z> 1.84) = 0.9418 + (1 - 0.9671) = 0.9747 6.2 a. P(-1.57<Z< 1.84) = 0.9671-0.0582 = 0.9089 b. P(Z < - 1.57) + P(Z> 1.84) = 0.0582 + 0.0329 = 0.0911 c. If P(Z> A) = 0.025, P(Z < A) = 0.975. A = + 1.96. d. If P(-A < Z < A) = 0.6826, P(Z < A) = 0.8413. So 68.26% of the area is captured between -A = - 1.00 and A = + 1.00. 6.4 a. P(Z> 1.08) = 1 - 0.8599 = 0.1401 b. P(Z<-0.21)=0.4168 c. P(- 1.96<Z<-021)=04168-00250=03918 d. P(Z> A) = 0.1587, P(Z < A) = 0.8413. A = + 1.00. 6.6 a. P(X> 43) = P(Z> - 1.75) = 1 - 0.0401 = 0.9599 b. P(X < 42) = P(Z < -2.00) - 0.0228 c. P(X < 4) = 0.05, Z =-1.645 = A - 50/4, A = 50 - 1.645(4) = 43.42 d. P(Xlower < X < Xupper) = 0.60 P(Z < - 0.84) = 0.20 and P(Z < 0.84) = 0.80 P(Xlower < X < Xupper) = 0.60 P(Z < - 0.84) = 0.20 and P(Z < 0.84) = 0.80 Z =-0.84 = Xlower -50/4 Z = +0.84 = Xupper -50/4 Xlower = 50 - 0.84(4) - 46.64 and Xupper = 50 + 0.84(4) = 53.36 6.8 a. P(34 < X < 50) = P(- 1.33 < Z < 0) = 0.4088 b. P(X < 30) + P(X > 60) = P(Z < - 1.67) + P(Z> 0.83) = 0.0475 + (1.0 - 0.7967) = 0.2508 c. P(X > A) = 0.80 P(Z<-0.84) = 0.20 A -50 7 = _0.84 = 12 d. The smaller standard deviation makes the Z-values larger. P(34 < X < 50) = P(- 1.60 < Z < 0) = 0.4452 P(X < 30) + P(X > 60) = P(Z < - 2.00) + P(Z > 1.00) = 0.0228 + (1.0 - 0.8413) = 0.1815 A = 50 - 0.84(10) = 41.6 thousand miles or 41,600 miles

8.38 a. n = Z² σ²/ e ² = 1.96² ・ 400²/50² = 245.86 using n = 246 b. n = Z² σ²/e² = 1.96² ・ 400²/25² = 983.41 using n = 984 Unit 6 9.3 Decision rule: Reject Ho if Z STAT < -1.96 or Z STAT > +1.96 9.4 Decision rule: Reject Ho if Z STAT < -2.58 or Z STAT > +2.58 9.12 H ០ : μ = 20 minutes. 20 minutes is adequate travel time between classes H1 : μ ≄ 20 minutes. 20 minutes is not adequate travel time between classes 9.25 a. H ០ : μ = 8.15 H1 : μ ≄ 8.15 Decision rule: reject H ០ if |tstat| > 2.0096 or p -value < 0.05 d.f. = 49 Test statistic: Tstat = X ̄ -μ/s/√n = 8.159 - 8.15/0.051/√50= 1.2478 p -value = 0.2180 Decision: Since |Tstat| < 2.0096 and the p-value of 0.2180 > 0.05, we do not reject H ០ - There is not enough evidence to conclude that the mean amount is different from 8.15 ounces. b. The p-value is 0.2180. If the population mean is indeed 8.15 ounces, the probability of obtaining a sample mean that is more than 0.009 ounces away from 8.15 ounces is 0.2180 9.34 a. H ០ : μ = 5.5 H1 : μ ≄ 5.5 H, : M + 5.5 Decision rule: Reject H ០ if |Tstat| > 2.680 d.f . = 49 Test statistic: |Tstat| = X ̄ -μ/√n = 5.5014 -5.5/0.1058 = 0.0935 Decision: Since |Tstat| < 2.680, do not reject H, there is not enough evidence to conclude that the mean amount of tea per bag is different from 5.5 grams b. X ̄ ± t ・ S/√n =5.5014 ± 2.6800 ・ 0.1058/√50 5.46<M < 5.54 You can conclude that the population mean amount of tea per bag is somewhere between 5.46 and 5.54 grams with 99% confidence. c. The conclusions are the same. Unit 7

3.47 13.1 a. When X = 0, the estimated expected value of Y is 2. b. For each increase in the value X by 1 unit, you can expect an increase by an estimated 5 units in the value of Y. c. Ÿ =2+5 X =2 + 5(3) = 17 13.6 a. B b. b0 = -13,130.6592, b1 = 2.4218. c. For each increase of $1,000 in tuition, the mean starting salary is predicted to increase by $2,421.80 d. $109,047.01 e. Starting salary seems higher for those schools that have a higher tuition 13.18 (graph needed) a. r 2 = 0.7665. 76.65% of the variation in starting salary can be explained by the variation in tuition b. S yx = 15,944.3807. c. Based on (a) and (b), the model should be very useful for predicting the starting salary 13.44 (graph needed) a. T stat = 10.7174 > 2.0301; p -value = 0.0000 < 0.05 reject Ho. There is evidence of a linear relationship between tuition and starting salary b. 1.963 ≤ β1 ≤ 2.8805 Unit 8 14.7 a. Ŷ =-330.675 + 1.764865X 1 - 0.13897 X 2 b. For a fixed quantity of remote hours, each additional unit in the total staff present is projected to lead to an average rise of 1.764865 standby hours. Similarly, with a constant total staff presence, every incremental unit in remote hours is anticipated to correlate with an average reduction of 0.13897 standby hours. c. The practical interpretation of b0 is negligible in this context since it signifies an estimation of mean standby hours when both total staff and remote hours are absent. d. Ŷ i =-330.675 + 1.764865(310) - 0.13897(400) = 160.845