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VashaNatasha [74]
3 years ago
7

Norman purchased $420 of stock in Carnival Cruise Lines last week. That stock is now worth $525. Use your knowledge of percent c

hange to determine the percentage by which the value of Norman's stock has increased. Be sure to show your work.
Mathematics
1 answer:
zysi [14]3 years ago
7 0

Answer:

The stock Norman purchased increased in its value by 25%.

Step-by-step explanation:

Original Value of the stock = $ 420

New(Current) value of the stock = $ 525

We have to find the percentage change in the value of stock. The formula to calculate the percentage change is:

\text{Percentage Change}=\frac{\text{New Value - Original Value}}{\text{Original Value}} \times 100\%

Substituting the given values into this formula results in:

\text{Percentage Change}=\frac{525-420}{420} \times 100\% \\\\ \text{Percentage Change}=\frac{105}{420} \times 100\%\\\\ \text{Percentage Change}=0.25 \times 100\% \\\\ \text{Percentage Change}=25 \%

A positive value of Percentage Change indicates a growth. This means that the stock Norman purchased increased in its value by 25%.

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Answer:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

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So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

\bar X_{1}=9.1 represent the mean for the sample 1

\bar X_{2}=8 represent the mean for the sample 2

s_{1}=1.9 represent the sample standard deviation for the sample 1

s_{2}=2.1 represent the sample standard deviation for the sample 2

n_{1}=40 sample size for the group 1

n_{2}=50 sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis:\mu_{1} = \mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

Statistical decision

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=40+50-2=88

Since is a bilateral test the p value would be:

p_v =2*P(t_{(88)}>2.604)=0.0108

So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

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