I had these two problems on my test but I didn't know how to do them:


1. A study was carried out to compare high-density flax-protein concentrations in adult men with sedentary jobs and manual labor. The sample data provided the following results:


 MeanDeviationSample size
Sedentary workers56.514.155
Manual workers51.313.550


Assume that the populations follow a normal distribution equal variances. Build a 95% confidence interval for the difference between the population means.


2. The average recovery time of a sample of 20 patients discharged from a general hospital is 7 days, with a standard deviation of 2 days. A sample of 24 patients discharged from a chronic disease hospital had an average recovery time of 36 days with a standard deviation of 10 days. Assume that the population follows a normal distribution with unequal variances and calculate the 95% confidence interval for the difference between the population means.



I appreciate the help since my teacher didn't answer me sadlaugh

chaconsara32  Mar 25, 2018

1+0 Answers


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. 




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.
\(\sigma_{\overline{x}_1-\overline{x}_2}\approx 2.6944\)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!

TheXSquaredFactor  Mar 31, 2018

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