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3

Test your sum of square $ s = 60.84 + \dots 51.84$. I get it as $s = 1003.20$. Then depending of the standard deviation type you get for the sample std. dev. $$\sigma_s = \sqrt{s/19} = 7.26636$$ or the population std. dev. $$\sigma_p = \sqrt{s/20} = 7.08237$$


4

There are two ways to calculate variance. First one is biased estimator and other is unbiased estimator of variance. In e-gadgets, probably estimator is unbiased estimator and that's what you are getting inconsistent result. Try dividing variance by $n-1$ , the number of data points instead of $n$ and then take square root. Good Luck!


0

ERROR BARS. As far as I can discover, the term 'error bar' can refer to almost any line that indicates a degree of uncertainty. In different disciplines and applications they can indicate the standard deviation (SD) of data, the standard error (often SD divided by square root of sample size), or confidence interval (90%, 95%, or 99%). See the very brief ...


0

If everyone is 5 feet tall in country A and 6 feet tall in country B, then each individual country has mean 5,6 height respectively, with zero variance. However if you want to compare the two, you need to know the population size. For example if there are $n_A,n_B$ people in both countries, then the mean height amongst both is: $m=\frac{5n_A+6n_B}{n_A+n_B}$. ...


0

One way to approach the problem might be to calculate the standard score as the number of standard deviations for the observation from the expected value: $$ \sigma = \sqrt{18000\times\frac{1}{6}\times(1 - \frac{1}{6})} = 50\\ z = \frac{3123-\frac{18000}{6}}{50} = 2.46 $$ If my understanding is correct, 2.46 is a high standard score which has less than 1% ...


0

Four speculative scenarios follow: Perhaps the context of the problem in the text or lectures could provide a clue which (if any) might be intended. 1) Uniform. Perhaps, somehow, "constant" is supposed to suggest a uniform distribution. To have mean 35, that would have to be $X \sim Unif(0, 70),$ so $P(X \le 45) = 45/70 = 0.6429.$ 2) Normal. Some yards may ...


2

When doing this test what we're interesting in is the distribution of our test statistic when the null hypothesis is true, which is binomial$(n = 18,000, p = 1/6)$. So the standard error need not be estimated from the data, it's known to be exactly $\sqrt{18,000 \times 1/6 \times 5/6}$, and this should be the value you use in your calculation. More ...


1

For the purpose of this answer I'll use the "% ground area covered by buildings". Since each of your data points corresponds to $1$ km by $1$ km tiles it actually doesn't "matter" if you use percentage or the raw data. Say one tile contains $.25 \text{km}^2$ of buildings. Then the percentage covering is $25\%$. In general if the raw data is $X$ then ...


1

If you know the entire population you're talking about, the formula with $n$ in the denominator will give you its true variance. However, if you don't know the entire population, but just have a limited amount of random samples from it, it is unlikely that your samples will have the full variation of the underlying population (that is, your random sample ...


0

I suppose ratio of MRR/Customer Churns as function of time is a good enough parameter for comparisons.


1

To avoid square roots for the moment, the definition of the sample variance is $$S^2 = \frac{\sum (X_i - \bar X)^2}{n-1}.$$ After a little algebra, the numerator can be written in several ways: $$(n-1)S^2 = \sum (X_i - \bar X)^2 = \sum X_i^2 - n\bar X^2 = \sum X_i^2 - \frac{(\sum X_i)^2}{n}.$$ Your expression (Σx^2-xbar^2)/(n-1) seems to use the second of ...


2

$\displaystyle \sqrt{\frac{n}{n-1}\left(\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2\right)}$ will give you $0.37$ with this data. This is an estimator of the standard deviation, not the standard error of the mean; that would be $\displaystyle \sqrt{\frac{1}{n-1}\left(\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2\right)}$.



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