Unleash Your Pooled Standard Deviation Potential with Expert Support and Comprehensive Practice

So I'm trying to test if there is statistically significant difference between the frequency of play of different recreational soccer players based on their stated reasons for play.
So for example: The average amount of times someone plays soccer per week when they state that exercise is their reason for play - compared to the frequency of someone who might say the social aspect is their reason. How might I set up such a hypothesis test.

this problem is from the wackerly, but i have a particular question, in this problem we have to estimate the variance with $S_p^2$ or can we use the information standard deviations that they gave us.
Here is the problem:
Two new drugs were given to patients with hypertension. The first drug lowered the blood pressure of 16 patients an average of 11 points, with a standard deviation of 6 points. The second drug lowered the blood pressure of 20 other patients an average of 12 points, with a standard deviation of 8 points. Determine a 95% confidence interval for the difference in the mean reductions in blood pressure, assuming that the measurements are normally distributed with equal variances.

I have 2 groups of people. I'm working with the data about their age. I know the means, the standard deviations and the number of people. I don't know the data of each person in the groups.
Group 1 :
Mean = 35 years old; SD = 14; n = 137 people
Group 2 :
Mean = 31 years old; SD = 11; n = 112 people
I want to combine those 2 groups to obtain a new mean and SD. It's easy for the mean, but is it possible for the SD? I do not know the distribution of those samples, and I can't assume those are normal distributions. Is there a formula for distributions that aren't necessarily normal?

I have 12 balls with mean weight of 500g, and S.D 25g. I have 5 balls with mean weight 200g and S.D 10g. If I combine all of them, is the mean just $\frac{8000}{17}$ i.e.
$8000=500\cdot <!--\; \cdot \; -->12+200\cdot <!--\; \cdot \; -->10$
and divided by 17 since there are 17 balls. What is the new S.D.? Is it just
$\sqrt{25\cdot <!--\; \cdot \; -->12+10\cdot <!--\; \cdot \; -->5}$

Divide the population of U.S. workers into two categories: those who have a job in the goods-producing industries, and those who have a job in the service-producing industries. Among all workers, sixteen percent have a job in the goods-producing industries. For these workers, the mean of annual earnings equals $\mathrm{\$<!--\; \$\; -->}42,419$, with a standard deviation of $\mathrm{\$<!--\; \$\; -->}35,572$. Among all workers, the remaining eighty-four percent have a job in the service-producing industries. For these workers, the mean of annual earnings equals $\mathrm{\$<!--\; \$\; -->}36,976$, with a standard deviation of $\mathrm{\$<!--\; \$\; -->}31,247$. Find the mean and standard deviation of annual earnings among all workers in the population.

Is this the correct answer:

Pop mean $(.16\ast <!--\; \ast \; -->42,419)+(.84\ast <!--\; \ast \; -->36,976)=37,846.88$
Pop StdDev $(.16\ast <!--\; \ast \; -->35,572)+(.84\ast <!--\; \ast \; -->31,247)=31,939$

Will the standard deviation of all data in three unequally-sized groups each with different means be the same if each group has the same standard deviation?

For example, I have a data set as follows:

Group 1, Makes up 30% of individuals, Mean = 12, SD = 3
Group 2, Makes up 50% of individuals, Mean = 17, SD = 3
Group 3, Makes up 20% of individuals, Mean = 11, SD = 3

I've been asked to calculate whether the SD for all individuals together is equal to 3, or larger than 3. My instinct is to say that it will be larger than 3 given that the the means are quite far apart, but can I be certain given just the information provided...? Any help here would be much appreciated.

Suppose we're studying the time it takes for a certain industrial process to complete. A recent study, which measured the time it took to complete 51 processes, gave a mean time of 2.396 minutes with standard deviation of 1.967 minutes. From past studies it has been observed that the standard deviation of the time it takes for the process to complete is 2.1 minutes. Calculate a confidence interval of 98% for the standard deviation of the time it takes to complete the process.

I don't understand why would they give me the value of the standard deviation of past processes, if I already have a value for $s^2$ (namely 1.967). By the way, I build the confidence interval by using the fact that $(n-<!--\; -\; -->1)S^2/{\mathrm{\sigma <!--\; \sigma \; -->}}^2$ follows a Chi squared distribution and then calculating the probability of ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$ being between the inferior and superior limits of the ${\mathrm{\chi <!--\; \chi \; -->}}^2$ distribution for which the area under the curve is equal to (1−0.98)/2. I do not know why is that past value there if the interval can be calculated using only the first value.

I remember that in class, the professor said that $S_p^2$ is the single best estimator of the variance of a single population in terms of the variance of another population, though I didn't understand that. I suspect I can use the pooled variance in order to calculate a more accurate confidence interval taking in account both values of the standard deviation that they give me?

And in general, what sense does it make to calculate the confidence interval for a variable whose value I already know?

At the printing company I work for, we have different materials that we print on; part numbers for those materials are assigned based on the unique supplier, material type, and material widths (so the same supplier could send us two different widths of the same material type, and they'd get different part numbers). With each part number, our supplier sends us a statistical summary of their quality control data - how many samples were taken (n), the average thickness of the samples, and the standard deviation. For the most common parts, we might get a few hundred different averages, Ns, and standard deviations. I'm trying to group those summaries together. Since the sample size is so high, I'm fine averaging the average thicknesses for each part number; the numbers are closely grouped enough that it shouldn't be a significant difference. My question is, how do I summarize the standard deviation? The quality tests are standardized and the instruments used to measure the thickness are constantly calibrated, so I have no reason to think that the measurements would be inconsistent from one shipment to the next.

Samples are taken from two different types of honey and the viscosity is measured.

Honey A:
Mean: 114.44
S.D : 0.62
Sample Size: 4

Honey B:
Mean: 114.93
S.D: 0.94
Sample Size: 6

Assuming normal distribution, test at 5% significance level whether there is a difference in the viscosity of the two types of honey?

Here's what I did:

I took my null hypothesis as $\mathrm{\mu <!--\; \mu \; -->}$B - $\mathrm{\mu <!--\; \mu \; -->}$A = 0 and alternative hypothesis as $\mathrm{\mu <!--\; \mu \; -->}$B - $\mathrm{\mu <!--\; \mu \; -->}$A $\ne <!--\; \ne \; -->$ 0

Then I did my calculations which were as following:

Test Statistic = (B -A ) - ($\mathrm{\mu <!--\; \mu \; -->}$B - $\mathrm{\mu <!--\; \mu \; -->}$A) / sqrt {(variance B / sample size B) + (variance A / sample size A)}

This gave me test statistic as = 0.49/0.49332 that is equal to 0.993

However the test statistic in the book solution is given as 0.91. What am I doing wrong?

I'm trying to solve following type of combined date given question. Imagined that there's two factories. They are X and Y. "X" factory has 10 workers and "Y" has 20 workers. These two factory's production respectively given as Σx^2=2950 and ΣY^2=5000. Moreover, "X" factory mean is 18 and "Y" factory mean is 15.
If the whole population is considered as a one. How to calculate the mean and standard deviation in a situation like this one? I'm not expecting the final answer, I rather would like to know the procedure that how to solve such a question.