A contractor has found that her cost for a certain construction job is subject to random variation. She believes its actual valu
e follows a continuous uniform distribution between $9000 and $11000 with a mean of $10,000. The cost to prepare her bid is $500. She is competing in a sealed bid competition with four other contractors. She believes that the bids submitted by the other contractors will vary triangularly with a minimum equal to her minimum cost, a most likely value equal to her 1.3 times her mean cost and a maximum equal to 2.5 times her mean cost and that their bids are independent of each other. She will win the competition if her bid is smaller than the bids of all four competitors. If she wins, her profit will be the difference between her bid and her cost minus the proposal preparation cost. If she loses her proposal preparation cost is lost and her profit is -$500. Simulate 1000 bids in which her bid amount is $14,000 and determine the distribution of her profit. Include the histogram in your response. Also determine the probability that she wins the competition and the probability that she loses money. Compare these results for bid amounts of $13000 and $15000
1 answer:
Answer:
Attached is the profit distribution plotted on the chart and also the detailed solution using excel
Step-by-step explanation:
To solve this problem we have to
- create a column for number of counts ( 1,2,3........1000) bids
- create a column for the cost to be incurred which is mostly dependent on the random number generated. the formula for that using excel is; 9000+ rand()*(11000-9000) for uniform distribution between the numbers
- Four(4) more columns are generated for bids of competitors by using the formula: 10000+rand()*(3*10000-10000) this because the bids that will be submitted by others bidders will vary uniformly between her mean cost and 3 times her mean cost
- Condition is checked to see if the lowest bid is. =IF(MIN(the 4 bids)>14000,1,0)
- Next the same process is carried out for 13000 and 15000
- The probability of winning is calculated in excel using this formula =Countif(value of step 4 for all the rows,1)/1000
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Answer:
a. As college debt increases current investment decreases.
b. Y= 68778.2406 - 1.9112X
Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.
c. There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.
d. Y= $59222.2406
e. R²= 0.9818
Step-by-step explanation:
Hello!
You have the information on a random sample of 20 individuals who graduated from college five years ago. The variables of interest are:
Y: Current investment of an individual that graduated from college 5 years ago.
X: Total debt of an individual when he graduated from college 5 years ago.
a)
To see the relationship between the information about the debt and the investment is it best to make a scatterplot with the sample information.
As you can see in the scatterplot (attachment) there is a negative relationship between the current investment and the debt after college, this means that the greater the debt these individuals had, the less they are currently investing.
The statement that best describes it is: As college debt increases current investment decreases.
b)
The population regression equation is Y= α + βX +Ei
To develope the regression equation you have to estimate alpha and beta:
a= Y[bar] -bX[bar]
a= 44248.55 - (-1.91)*12829.70
a= 68778.2406
b=
b=
b= -1.9112
∑X= 256594
∑X²= 4515520748
∑Y= 884971
∑Y²= 43710429303
∑XY= 9014653088
n= 20
Means:
Y[bar]= ∑Y/n= 884971/20= 44248.55
X[bar]= ∑X/n= 256594/20= 12829.70
The estimated regression equation is:
Y= 68778.2406 - 1.9112X
Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.
c)
The hypotheses to test if there is a linear regression between the two variables are two tailed:
H₀: β = 0
H₁: β ≠ 0
α: 0.01
To make this test you can use either a Student t or the Snedecor's F (ANOVA)
Using t=<u> b - β </u>=<u> -1.91 - 0 </u>= -31.83
Sb 0.06
The critical region and the p-value for this test are two tailed.
The p-value is: 0.0001
The p-value is less than the level of signification, the decision is to reject the null hypothesis.
Using the
The rejection region using the ANOVA is one-tailed to the right, and so is the p-value.
The p-value is: 0.0001
Using this approach, the decision is also to reject the null hypothesis.
The conclusion is that at a 1% significance level, there is a linear regression between the current investment and the college debt.
The correct statement is:
There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.
d)
To predict what value will take Y to a given value of X you have to replace it in the estimated regression equation.
Y/X=$5000
Y= 68778.2406 - 1.9112*5000
Y= $59222.2406
The current investment of an individual that had a $5000 college debt is $59222.2406.
e)
To estimate the proportion of variation of the dependent variable that is explained/ given by the independent variable you have to calculate the coefficient of determination R².
R²= 0.9818
This means that 98.18% of the variability of the current investments are explained by the college debt at graduation under the estimated regression model: Y= 68778.2406 - 1.9112X
I hope it helps!
