Можно ли совершить кругосветное путешествие по меридиану на север? - коротко
Кругосветное путешествие строго по меридиану на север невозможно, так как все меридианы сходятся на Северном полюсе, и движение вдоль них приведёт только к достижению этой точки.
Можно ли совершить кругосветное путешествие по меридиану на север? - развернуто
Кругосветное путешествие по меридиану на север невозможно в буквальном смысле, так как меридиан — это воображаемая линия, соединяющая Северный и Южный полюса. Двигаясь строго по меридиану, можно достичь Северного полюса, но дальнейшее движение в том же направлении приведет к изменению курса на юг. Таким образом, кругосветное путешествие предполагает движение вокруг Земли по параллелям, а не по меридианам.
Если рассматривать идею путешествия, которое охватывает весь земной шар, включая Северный полюс, то это возможно, но потребует сложного маршрута. Например, можно начать движение от экватора, двигаться строго на север до Северного полюса, затем продолжить путь на юг по другому меридиану, достигнув противоп# 2.1 - Sample Size for Estimating Population Mean and Total
How large is a sample size that is large enough for estimating the population mean?
If (\hat{\theta}) is an unbiased, normally distributed estimator of (\theta), then
(\dfrac{\hat{\theta}-\theta}{\sqrt{Var(\hat{\theta})}} \sim N(0,1))
Then (P\left(\dfrac{|\hat{\theta}-\theta|}{\sqrt{Var(\hat{\theta})}} > z_{\alpha/2} \right)=\alpha )
( P\left(|\hat{\theta}-\theta|>z_{\alpha/2} \cdot \sqrt{Var(\hat{\theta})} \right)= \alpha )
Note! because we know that (\hat{\theta}) is normal, we can thus use the z distribution.
And, if we specify this (\alpha) we can then try to find out the sample size large enough to achieve the goal of your experiment.
So, we need to ask, "What is the goal of your experiment?" This is perhaps the most important question asked as a part of your experiment.
Example: What if we were interested in estimating the average weight of Penn State male students? How many samples should we plan on taking? We want to estimate this mean. What do we need to consider?
- The variability of the data and the measure that you are estimating is your first concern. This directly affects how many samples you will need.
- The second thing that you need to think about is the type of conclusion that you would like to report. That is, you need to specify the (1 - \alpha) value that you are happy with.
- How accurate (precision) do you want this estimate to be? You thus need to specify the margin of error.
Now, if we specify (1-\alpha), the margin of error d (also can be viewed as the half-width of the ((1 - \alpha))100% CI), we can solve for the sample size such that the CI has the specified margin of error.
For estimating the population mean, the equation becomes:
(P\left(|\bar{x}-\mu|>z_{\alpha/2} \cdot \sqrt{\dfrac{N-n}{N}\cdot \dfrac{\sigma^2}{n}}\right)=\alpha)
(z_{\alpha/2}\sqrt{\dfrac{N-n}{N}\cdot \dfrac{\sigma^2}{n}}=d)
(n=\dfrac{1}{\dfrac{ d^2}{ z^2_{\alpha/2}\cdot \sigma^2}+\dfrac{1}{N}})
Can we now use this formula to estimate the sample size?
What is the weak point of this formula? The weak point is the estimate of the population variance used. We do not know what this is!
Similarly, for estimating the population total (\tau), here is the formula:
(P\left(|\hat{\tau}-\tau|>z_{\alpha/2} \cdot \sqrt{N(N-n)\dfrac{\sigma^2}{n}} \right)=\alpha)
(z_{\alpha/2}\sqrt{N(N-n)\dfrac{\sigma^2}{n}}=d)
(n=\dfrac{1}{\dfrac{ d^2}{ N^2 \cdot z^2_{\alpha/2}\cdot \sigma^2}+\dfrac{1}{N}})
Example 2-1: Beetles - Sample size Section
What sample size is needed to estimate the population total, (\tau), to within d = 1000 with a 95% CI?
Now, let's begin plugging what we know into the formula. We know N = 100, (\alpha) = 0.05. Do we know (\sigma^2)? No, but we can estimate (\sigma^2) by (s^2) = 1932.657.
How many should we sample? Let's calculate this out:
(n=\dfrac{1}{\dfrac{ d^2}{ N^2 \cdot z^2_{\alpha/2}\cdot \sigma^2}+\dfrac{1}{N}})
(n=\dfrac{1}{\dfrac{ (1000)^2}{ (100)^2 \cdot (1.96)^2 \cdot 1932.657}+\dfrac{1}{100}}=42.610)
We will always round this up, therefore, we will sample 43 of the 100 plots.
Note! If we ignore the finite population correction adjustment then,
(n=\dfrac{N^2 \cdot z^2_{\alpha/2} \cdot \sigma^2}{d^2})
(n=\dfrac{(100)^2 \cdot (1.96)^2 \cdot 1932.657}{(1000)^2}=74.44)
round up to 75.
This is a much larger sample because we have not used the extra information that we know about the population size and have applied the finite population correction to obtain a more efficient (smaller) sample size estimate.
Try it!
What sample size is needed to estimate the population total, (\tau), to within d = 1000 with a 95% CI assuming the samples are taken with replacement?
(n=\dfrac{z^2_{\alpha/2} \cdot N^2 \cdot \sigma^2}{d^2})
(n=\dfrac{(1.96)^2 \cdot (100)^2 \cdot 1932.657}{(1000)^2}=74.44)
Round up to 75.
Note that if we do not use the finite population correction in the first example, we get the same sample size of 75. This is because the population size N is large compared to the sample size n. When the population size is large, the finite population correction factor does not change the sample size by very much. When the population size is small, however, the finite population correction factor will make a larger difference in the sample size estimate. For example, if the population size is 100 and the sample size is 25 then the correction factor is approximately 0.87. This means that the sample size can be reduced to 87% of the original estimate.