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1 year ago

Distinguish between a correlation and an autocorrelation. how are these measures similar? how are they different?

1 answer:

3 0

A correlation is a causative association between two or more variables, whereas an autocorrelation is a correlation of variables in successive ranges.

<h3>What is correlation?</h3>

A correlation is defined as a causative association between two or more variables (different variables), which can be used to make predictions about a given outcome.

The correlation coefficient enables the estimation of this association between different variables.

Moreover, an autocorrelation is a special type of correlation between variables that can be found in successive ranges (e.g. range time intervals).

In conclusion, a correlation is a causative association between two or more variables, whereas an autocorrelation is a correlation of variables in successive ranges.

Learn more about correlation here:

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jok3333 [9.3K]

Answer:

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Step-by-step explanation:

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Which of the following is equivalent to 3^4?<br> 12<br> 81<br> 07<br> 64

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Mr.king used 16 gallons of gas in 5 days. How much gallons of gas did gas mr king use per day

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He would use 3.2 gallons a day. All you need to do is use the calculator app and divide 16 and 5 to get 3.2. Hope this helps!!

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3 years ago

The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp

vovikov84 [41]

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, $\sigma$ = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15

(a) Let Null Hypothesis, $H_0$ : Mean Diameter, $\mu$ = 8.2500 cm

Alternate Hypothesis, $H_1$ : Mean Diameter, $\mu$ $\neq$ 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

$\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ follows Standard Normal distribution

After putting each value, Test Statistics = $\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }$ = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < $1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

P( $-Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $-\mu$ < $1.96\frac{\sigma}{\sqrt{n} }-Xbar$ ) = 0.95

P( $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

So, 95% confidence interval for $\mu$ = [ $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ , $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ]

= [ $8.2535-1.96\frac{0.0020}{\sqrt{15} }$ , $8.2535+1.96\frac{0.0020}{\sqrt{15} }$ ]

= [8.2525 , 8.2545]

Here $\mu$ = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

= [8.2525 , 8.2545] .

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ANTONII [103]

Answer:

I'm guessing the right answer should be B

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3 years ago