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Nata [24]
1 year ago
9

If the coefficient of determination is 0.233, what percentage of the variation in the data about the regression line is explaine

d?5.43%76.7%23.3%46.6%
Mathematics
1 answer:
ddd [48]1 year ago
3 0

We need the coefficient of determination definition

The coefficient of determination (R²) is a number between 0 and 1 that measures how well a statistical model predicts an outcome. You can interpret the R² as the proportion of variation in the dependent variable that is predicted by the statistical model

So if we have a coefficient of determination of 0.233 we multiply by 100 to get the percentage

Answer: 23.3%

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An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume
Lynna [10]

Answer:

a) 21-2.58\frac{52}{\sqrt{350}}=13.83  

21+2.58\frac{52}{\sqrt{350}}=28.17  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

b) ME=2.58\frac{52}{\sqrt{350}}=7.17

c) ME=2.58\frac{52}{\sqrt{100}}=13.42

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

\bar X=21 represent the sample mean  

\mu population mean (variable of interest)  

\sigma=52 represent the population standard deviation  

n=350 represent the sample size  

99% confidence interval  

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58  

Now we have everything in order to replace into formula (1):  

21-2.58\frac{52}{\sqrt{350}}=13.83  

21+2.58\frac{52}{\sqrt{350}}=28.17  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

Part b

The margin of error is given by:

ME=2.58\frac{52}{\sqrt{350}}=7.17

Part c

The margin of error is given by:

ME=2.58\frac{52}{\sqrt{100}}=13.42

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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Answer:

I think true I'm not sure

4 0
3 years ago
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The answer is ob.24
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Answer:

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Step-by-step explanation:

Given that,

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Hence, this is the required solution.

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Answer:

Step-by-step explanation:

Alina=2x---eq(1)

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So,we get

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Therefore Alina got 60 Rs while Asim got 150 Rs

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3 years ago
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