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Vitek1552 [10]
3 years ago
5

Use the model below to estimate the average annual growth rate of a certain country's population for 1950, 1988, and 2010, where

x is the number of years after 1900.
Y= -0.0000084x^3 + 0.00211x^2 - 0.205x + 8.423

The estimated average annual growth rate of the country's population for 1950 is?
Mathematics
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer:

The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.

Step-by-step explanation:

The given equation is

Y=-0.0000084x^3+0.00211x^2-0.205x+8.423

Where Y is the annual growth rate of  a certain country's population and x is the number of years after 1900.

Difference between 1950 and 1900 is 50.

Put x=50 in the given equation.

Y=-0.0000084(50)^3+0.00211(50)^2-0.205(50)+8.423

Y=2.398

Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.

Difference between 1988 and 1900 is 88.

Put x=88 in the given equation.

Y=-0.0000084(88)^3+0.00211(88)^2-0.205(88)+8.423

Y=0.9984752\approx 0.9985

Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.

Difference between 2010 and 1900 is 110.

Put x=110 in the given equation.

Y=-0.0000084(110)^3+0.00211(110)^2-0.205(110)+8.423

Y=0.2236

Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.

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Answer:

z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219  

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If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes.    

Step-by-step explanation:

Data given and notation  

\bar X=9.25 represent the sample mean  

\sigma=2 represent the population standard deviation

n=35 sample size  

\mu_o =10 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z 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 10 minutes, the system of hypothesis would be:  

Null hypothesis:\mu \geq 10  

Alternative hypothesis:\mu < 10  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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:  

z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219  

P-value  

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

p_v =P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes.    

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