TheXSquaredFactor

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BenutzernameTheXSquaredFactor
Punkte1845
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 #1
avatar+1845 
+1

We have two pieces of data from a study from separate groups.

 

In order to create the baseline, it is necessary to find the difference of the individual means of both data.

 

\(\text{Sedentary: }\overline{x}_1=56.5, SD_1=14.1,s_1=55\\ \text{Manual: }\overline{x}_2=51.3,SD_2=13.5,s_2=50\)

 

SD = sample standard deviation

s1 and s2 are the individual values for sampling size.

 

Since the question asks for the difference between the population means,  one should do that in order to create a baseline. 

 

\(\mu_{\overline{x}_1-\overline{x}_2}=\overline{x}_1-\overline{x}_2=56.5-51.3=5.2\) 

 

This is the exact middle of the data. In order to establish a 95% confidence level of the difference, we will have to find the number of standard deviations from the current \(\mu_{\overline{x}_1-\overline{x}_2}\) . We can utilize the empirical rule (sometimes referred to the 68-95-99.7% rule) for this. An image will probably do the best explaining.

 

Source: http://simulation-math.com/_Statistics/EmpiricalRule.png

 

This image indicates to me that 95% of the data falls within 2 standard deviations from the mean. Because a bell-shaped distribution is symmetric in nature, we know that, from the mean, 95% will fall between \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) . By calculating this information, there should be a 95% chance that 5.2 (the \(\mu_{\overline{x}_1-\overline{x}_2}\)) lies within \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) of the mean. Now, we must find that standard deviation. Let's use some algebra manipulation to find a formula for this.

 

\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\sigma_{\overline{x}_1}^2+\sigma_{\overline{x}_2}^2\) In other words, the variance of the means of the distribution equals the sum of the variance of each sampling distribution. We can rewrite this equation further because the variance of each sampling distribution also equals the variance of the population distribution divided by the sampling size. 
\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\frac{\sigma_{x_1}^2}{s_1}+\frac{\sigma_{x_2}^2}{s_2}\) There is only one problem: We do not know the sample variances. However, we can approximate by using the values from the sample standard deviations instead.
\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\frac{SD_{1}^2}{s_1}+\frac{SD_{2}^2}{s_2}\) Now, solve for sigma by taking the square root of both sides.
\(\sigma_{\overline{x}_1-\overline{x}_2}=\sqrt{\frac{SD_{1}^2}{s_1}+\frac{SD_{2}^2}{s_2}}\) Now, plug in the values for variables in the equation. Normally, I would rationalize the denominators, but this is a calculator's job anyway, so it does not really matter.
\(\sigma_{\overline{x}_1-\overline{x}_2}=\sqrt{\frac{14.1^2}{55}+\frac{13.5^2}{50}}\)  
\(\sigma_{\overline{x}_1-\overline{x}_2}\approx 2.6944\) This is the variance calculated to the nearest ten-thousandth place. Since 95% is two standard deviations away, we need to double this answer.
\(2\sigma_{\overline{x}_1-\overline{x}_2}\approx 2*2.6944\approx5.3888\)  


Remember this statement from earlier? I said there should be a 95% chance that 5.2 (the \(\mu_{\overline{x}_1-\overline{x}_2}\)) lies within \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) of the mean. We can derive the interval with this. Therefore, 95% of the data lies within \(5.2\pm5.3888\) .

 

We are, therefore, 95% confident that the true mean lies between \(-0.1888\) and \(10.588\) .

 

I encourage you to try the next one! See how you do!

 #1
avatar+1845 
+2

The domain is the set of all x-values that outputs a real y-value for a given equation.

 

\(f(x)=\sqrt[3]{x}+2\)

 

In this case, this particular function will accept any input and produce a real output. Because this is a cube-root function, the cube-root can produce any result of reals, so \(x\in \mathbb{R}\text{ or }(-\infty,+\infty)\) . Only one answer choice fits this criterion, so the 3rd answer must be the correct one. Let's consider the next function!

 

\(f(x)=-4\sqrt{x}\)

 

When a variable is placed underneath the square root (known as the radicand), there is a certain domain restriction that must be taken into account: The radicand must be nonnegative; inserting a negative number as the radicand will result in a non-real output since no real number multiplied by itself can equal a negative number. Because of this fact, the domain is \(x\geq 0\text{ or }[0,+\infty)\) . Unfortunately, this is not enough to eliminate every answer choice because two answer choices include this as a possibility. Therefore, we will have to figure out the range.

 

The range is the set of all outputs that the function can produce

 

Because of the definition of the square root operation, it is implied that this is a case of the principal root. Disregarding the rest of the function, \(\sqrt{x}\) will always output a positive value or zero. Multiplying any positive number or zero by a negative number will always result in a negative number or zero, so the output is any real non-positive number or \(f(x)\leq 0\text{ or }(-\infty,0]\)

 

With this knowledge, only the third answer choice remains.