Broadway Caf Final Project :
Porters five forces of competitive position analysis:
Buyer Power:
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
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:
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
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
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
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
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
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
217 19 18 20 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
Explanations: 1. Random variable X has 4 possible values to take in this case. X1 X2 X3 X4 2. P(X1) = 0.28 indicates there is a 0.28 probability (28%) for the random variable X to take the possible value X1=5 Condition 1 Each P(Xi) should be between 0 and
A. 1. n=12 2. p=.10 3. let x= # of people in the survey who prefer the cookie x is a RV with possible values 4. the probability that there are exactly x successes among the n=12 trials is P(X)=N!/X!(N-X)! * p^X(1-p)^N-X P(X=4)=P(4)=12!/4!(12-4)!(.10)^4(1-
X=# of times an event will occur in a time period X is a RV E=natural base=2.71828 227 21 23. 1. let x=number of hurricanes to hit usa in a year. 2. The mean number of hurricanes to hit usa in the year is u=0.6 3. the probability that there are X hurrican
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
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<
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
(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
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
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.
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)