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3 years ago

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Imagine that the average weight of a total of 500 girls in a high school is 35kg. Tom randomly sampled 10 girls and measured the

ir weight. And then he repeated this procedure for three times. The means and standard deviations are listed as following. Which sample estimate shows the least sample variability?

1 answer:

Naddik [55]3 years ago

4 0

Answer:

The sample 2 has a lowest value of SE corresponding to the least sample variability.

Step-by-step explanation:

As the value of the sample means and standard deviations are not given, as similar question is found online from which the values of data is follows

The data is as attached with the solution. From this data

Sample 1 has a mean of 34 and a SE of 5

Sample 2 has a mean of 30 and a SE of 2

Sample 3 has a mean of 26 and a SE of 3

Sample 4 has a mean of 38 and a SE of 5

As per the measure of the sample variability is linked with the value of SE or standard error. Which is lowest in the case of sample 2 .

So the sample 2 has a lowest value of SE corresponding to the least sample variability.

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An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1-ounce p

ludmilkaskok [199]

Answer:

a) $y=-0.317 x +46.02$

b) Figure attached

c) $S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

Step-by-step explanation:

We assume that th data is this one:

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

a) Find the least-squares line appropriate for this data.

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720$

$\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324$

$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

$m=-\frac{1900}{6000}=-0.317$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60$

$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

And we can find the intercept using this:

$b=\bar y -m \bar x=27-(-0.317*60)=46.02$

So the line would be given by:

$y=-0.317 x +46.02$

b) Plot the points and graph the line as a check on your calculations.

For this case we can use excel and we got the figure attached as the result.

c) Calculate S^2

In oder to calculate S^2 we need to calculate the MSE, or the mean square error. And is given by this formula:

$MSE=\frac{SSE}{df_{E}}$

The degred of freedom for the error are given by:

$df_{E}=n-2=12-2=10$

We can calculate:

$S_{y}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=9540-\frac{324^2}{12}=792$

And now we can calculate the sum of squares for the regression given by:

$SSR=\frac{S^2_{xy}}{S_{xx}}=\frac{(-1900)^2}{6000}=601.67$

We have that SST= SSR+SSE, and then SSE=SST-SSR= 792-601.67=190.33[/tex]

So then :

$S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

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Answer:

Step-by-step explanation:

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Howdoyouwrite 2/4 as a percentage? write your answer using a percent sign %?

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Hakeem can make 7 free throws in 2 minutes how many would he make in 18 minutes if you continue at the same rate

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Answer:

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Step-by-step explanation:

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