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ivolga24 [154]
3 years ago
5

Which is an example of a high-risk investment?

Mathematics
2 answers:
solong [7]3 years ago
8 0

<u>Option (a) is correct. Stock in a start-up company is an example of high-risk investments. </u>

Further explanation:

High-risk investment:

The high-risk investment refers to the investment where the chances of capital loss are very high. The high-risk investment tends to provide a high rate of return on the cost of a high level of risk. The investment in the penny stock is an example of high-risk investment.  

Justification for the correct and incorrect option:

a.

Stock in a start-up company: This is the correct option because the start-up company is highly risky as it does not have an established formation and tends to lose the capital easily.

b.

Bond: This is an incorrect option because the bond is the loan and the bond has the preferential right to receive the payment of the interest and capital.

c.

CDs from an insured bank: This is an incorrect option because the security is backed up by the insured bank.

d.

401(k): This is an incorrect option because it is a retirement plan.

Thus, option (a) is correct. Stock in a start-up company is an example of high-risk investments.

Learn more:

  1. Learn more about consumer protection law  brainly.com/question/1862829
  2. Learn more about the laws related to the sole proprietor  brainly.com/question/5742225
  3. Learn more about the property damage law  brainly.com/question/4725726

Answer details:

Grade: Senior School

Subject: Business studies

Chapter: Investments

Keywords: high-level risk, bond, stock, shares, 401(k), capital protection, CDs, insured bank.

rewona [7]3 years ago
4 0

stock in a start-up company


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t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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