Lesson 8 DQs
When she saw the person, which was hurt much worse than her, in the hospital walk
for the first time, and their fa
Lesson 6 DQs
Ranchers used their logo to brand their catle. Now it holds the reputation of
Lesson 7 DQs
Its not as safe as a traditional
A fragmented industry is one composed of a:
large number of small and medium-sized firms.
large companies and their subsidiaries.
companies that operate in different locations across the world.
a small number of singl
Focus strategy can be defined as:
the strategy of closing down one or more business units in order to minimize the losses.
the strategy a company uses when it decides to ignore the different needs of different market segmen
One of the primary roles of research and development in achieving superior efficiency is:
instituting training programs to build skills.
designing products for ease of manufacture.
facilitating cooperation among function
Broadway Caf Final Project :
Porters five forces of competitive position analysis:
There are many coffee shops in the market, which makes higher number in competitors for The
Broadway Caf and gives more choices / substitutes for customers. Th
Principles of Marketing Chapter 13/14 F10 Online Chapter 13
Name: Aaron Rumley
1. What is a marketing channel? Marketing Channel: Set of interdependent organizations that eases the transfer of ownership as products move from producer to business user or c
Employees in a large accounting firm claim that the mean salary of the firms accountants is less then that of its competitors, which is $45,000. A random sample of 30 of the firms accountants has a mean salary of $43,500 with a std dev of $5200. at a=0.05
1. Ux=U Ux Is the mean of x_ U is the mean of the population 2. Ox=O/SQRT of N Ox is the st dev of X_ O is the std dev of the population N is the sample size. 280 26 28 25. 1. The pop. Number cfw_annual salaries of all plumbers 2. The mean of the populati
3rd test, ch 6 & 7 case A, o=0.10 Zo/z=1.645 case B, o=0.05 Zo/z=1.96 case C, o=0.01 Zo/z=2.575 318 38 46 41. Known that: 1. The population =cfw_repair costs of all broken washing machines The population mean U (i.e., the average repair cost of all broken
54. 4 16. (0.816, 0.843) >95% CI (0.812, 0.848) > 99% CI 20. (0.090, 0.129) >95% CI (0.084, 0.135) >99% CI Ch 6, topics covered in ch 7 Ho >=,<=, = Ha >, <, Not equal to. 393 41. 1 Population (life spans of all light bulbs produced by the factory) mean, u
Ch 7 testing for 1 pop 1. Testing for u 2. Testing for p LTT Ho:p1-p2>=Po Ha: p1-p2<Po Z
Ch 8 testing for 2 pop 1 Testing for u1 vs u2 2. Testing for p1 vs p2
If Z<-Za, reject at the a level of significance p^1, p^2 Two tailed test Ho: p1-p2=Po Ha:
12 Ho P1-P2>= 0 Ha P1-P2<0 2 1.5 z=-2.177 3 rr z=-2.177 -Za=Z0.05=-1.645 So, Z If Z< -Za, reject Ho at a=0.05 level Calculate TS T= R / SQRT(1-r^2/N-2) If T<-Ta/2,n-2 or T>+Ta/2,n-2, reject Ho at a level of sig. T=1.9495 B1 =index(linest(C2:C6,B2:B6),1) B
1. total tirals is 10 2. probability of csuccess is 21% 3. Let x = #of successes Part A. P(X=3)=? P(X=3)=P(3) 10! / 3!(10-3)! (.21)^3 (1-.21)^10-3 n=10 x=3 p=chance (.21) 10! = 1*2*3*4 10 3!=1*2*3 0!=1 p(e1)=1-p(e) P(E)=1-P(E1) P(X<=3)=P(0)+P(1)+P(2)+P(3)
Its known that: 1. Let X = height of a randomly selected man X is a normal RV 2. The mean of X is Ux=69.6 inches 3. The standard Deviation of X is Ox=3 inches A) P(X<66)=? P(X<66)=P(X-Ux/Ox < 66-Ux/Ox) =P(Z<66-69-6/3.0) P(Z<-3.6/3) P(Z<-1.20) -1.2=-1.2-0.
Confidence level (confidence coefficient) 1-a Takes 3 values, 0.01, 0.05, 0.1 Za/2 case A, o=0.10 Zo/z=1.645 case B, o=0.05 Zo/z=1.96 case C, o=0.01 Zo/z=2.575 Known that: 1. The population =cfw_repair costs of all broken washing machines The population m
(1.96)^2 0.5(1-0.5) / 0.03^2 n=1067.111=1068 Part B. ^p=0.26 (Za/2)^2 ^p(1-^p) / E^2 1.96^2 (0.21)(0.7) / 0.03^2 =821.25 o=822 Confidence intervals hypothesis testing correlation & linear regression Ch6. ci for u, p Ch7. testing for u, p Hypothesis testin
Binomial Distribution 1. Main Concern: x=# of successes among N Trials x=RV with all possible values (0,1,2) P(X)= N!/X!(n-x)! p^x(1-p)^n-x Geometric Distribution 1. Main Concern: X=Trial number on which the first success occurs X=RV with all possible val
257 17 C P(X>10)=? =P(X-Ux/Ox > 10-Ux/Ox) =P(Z>10-7/1) =P(Z>3.00)=The marked area =1-the marked area +3.00=+3.0+0.00 1-P(Z<3.00)=1-0.9987=0.0013 There is a 0.0013 probability that the randomly picked guy spends more than 10 hours a week on his pc. A) P(X<
P(X) = N!/X!(N-X)! * pX(1-P) ^ N-X X= the trial number on which the first defective PC is found 227 20 1. the possibility of asuccess of a single call is p=0.19 2. Let X=the trial number on which the first sale occurs 3. the probability that the first sal