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I need the following attachement of answers to be put into excel spreadheet with the correct formulas and graph. This needs to be done within the next 3 hours.
Probability
1. Suppose that the mean of the annual return for common stocks from 2000 to 2012 was 7.2%, and the standard deviation of the annual return was 31.2%. Suppose also that during the same 12-year time span, the mean of the annual return for long-term government bonds was 1.6%, and the standard deviation was 7.0%.
T
he distributions of annual returns for both common stocks and long-term government bonds are bell-shaped and approximately symmetric in this scenario. Assume that these distributions are distributed as normal random variables with the means and standard deviations given previously.
1. Find the probability that the return for common stocks will be greater than 3.5%.
2. Find the probability that the return for common stocks will be greater than 10%.
Hint: There are many ways to attack this problem in the HW. If you would like the normal distribution table so you can draw the pictures (my preferred way of learning) then I suggest you bookmark this site:
http://www.statsoft.com/textbook/sttable.html
common stock
mean= 7.25
SD= 31.25
1. Find the probability that the return for common stocks will be greater than 3.5%.
Z= (3.5 -7.25) /31.25= -0.12
Pr(X ≥ 3.5) = Pr(Z ≥ (3.5 – 7.25)/31.25 ) = Pr(Z ≥ -0.12)
= 1 – 0.4522 = 0.5478
2. Find the probability that the return for common stocks will be greater than 10%.
Z= (10-7.25 ) /31.25
P(z>10) = 0.088
Therefore, Pr(X ≥ 10) = Pr(Z ≥ (10 – 7.25)/31.25 ) = Pr(Z ≥ 0.088)
= 1 – 0.5351 = 0.4649
Confidence Interval Estimation
2. Compute a 95% confidence interval for the population mean, based on the sample 50, 54, 55, 51, 52, 51, 54, 52, 56, and 53. Change the last number from 53 to 91 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval.
Mean= 52.8
SD = 1.93
N= 10
At 95% confidence interval, z= 1.96
Mean ± z x σ/√n
52.8 ± 1.96
x1.93 /√10
52.8 ± 1.20
Confidence interval is from ( 51.6 to 54)
With change of 53 to 91
Mean= 56.6
SD = 12.24
N= 10
56.6 ± 1.96 x 12.24 /√10
56.6 ± 7.59
Confidence interval is from ( 49.01 to 64.19)
T
he effect of an outlier or extreme value on the confidence interval is that it increases the range of interval.
Hypothesis Testing
3. The management of XYZ Corporation is considering relocating the corporate office to a new location outside the Capital Beltway. Management is concerned that the commute times of the employees to the new office might be too long.
The company decides to survey a sample of employees at other companies in the same office park to see how long these employees are commuting to the office.
A sample of 18 employees indicated that the employees are commuting X (bar) = 40 minutes and s = 5 minutes.
a. Using the 0.05 level of significance, is there evidence that the population mean is above 35 minutes?
Ho= µ = 35
H1: µ >35
Level of significance = 0.05
Mean = 40 minutes
SD= 5 minutes
N= 18
Z= (35 – 40)/5 /√18 = -4.24
Critical value of z at 0.05 level for one –tailed test is 1.645.
As the calculated value of z is more than the critical value, we reject the null hypothesis.
There is sufficient evidence that the population mean is above 35 minutes.
b. What is your answer in (a) if X (bar) = 42 minutes and s = 20 minutes?
Ho= µ = 35
H1: µ >35
Level of significance = 0.05
Mean = 42 minutes
SD= 20 minutes
N= 18
Z= (35 – 42)/20 /√18 = -1.48
Critical value of z at 0.05 level for one –tailed test is 1.645.
As the calculated value of z is less than the critical value, we fail to reject the null hypothesis.
There is no evidence that the population mean is above 35 minutes.
c. Look at your answers for a and b above and discuss what you can learn from the results about the effect of a large standard deviation.
As the standard deviation increases, value of z decreases. The increase in the value of standard deviation causes the acceptance of the z –value and we accept the null hypothesis.
4. BestCoffeeInTown is concerned that the mean wait time of customers for a table is not greater than 7 minutes. It can be assumed that the population standard deviation is 2.8 minutes based on past experience. A sample of 300 customers is selected and the sample mean is 7.6 minutes. Using a level of significance of .01, is there evidence that the population mean wait time is greater than 7 minutes? Fully explain your answer.
Population SD = 2.8 minutes
Sample Size = 300
Sample mean= 7.6 minutes
Level of significance = 0.01
Population mean = 7 or more than 7
H0; µ = 7
H1: µ >7
It is a one tailed test
Z= ( 7-7.6)/2.8 =-0.2143
Therefore, Pr(X ≥ 7) = Pr(Z ≥ (7 – 7.6)/2.8 ) = Pr(Z ≥ -0.2143)
= 1 – 0.4152 = 0.5848
Critical value of z at 0.01 for 1 tail test is 2.33.
As the calculated value z is less than the critical value of z, we fail to reject the null hypothesis.
At level of significance of .01, there is no evidence that the population mean wait time is greater than 7 minutes.
