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UkoKoshka [18]
4 years ago
10

Consider two x distributions corresponding to the same x distribution. The first x distribution is based on samples of size n =

100 and the second is based on samples of size n = 225. Which x distribution has the smaller standard error? The distribution with n = 100 will have a smaller standard error. The distribution with n = 225 will have a smaller standard error. Explain your answer. Since σx = σ2/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ2. Since σx = σ/n, dividing by 100 will result in a small standard error regardless of the value of σ. Since σx = σ/n, dividing by 225 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ. Since σx = σ2/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ2.
Mathematics
1 answer:
Tanya [424]4 years ago
8 0

Answer:

The distribution with n = 225 will give a smaller standard error.

Since sigma x = sigma/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of sigma.

Step-by-step explanation:

Standard error is given by standard deviation (sigma) divided by square root of sample size (√n).

The distribution with n = 225 would give a smaller standard error because the square root of 225 is 15. The inverse of 15 multiplied by sigma is approximately 0.07sigma which is smaller compared to the distribution n = 100. Square of 100 is 10, inverse of 10 multiplied by sigma is 0.1sigma.

0.07sigma is smaller than 0.1sigma

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Find the following Euler Totients using Euler’s Theorem, as explained on p.409 of the text (10 points each): a.ϕ(13) b.ϕ(81) c.ϕ
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(a) \varphi(13)=12 since 13 is prime.

(b) 81=3^4, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so \varphi(81)=54.

(c) 100=2^2\cdot5^2; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with \varphi(100)=100-60=40.

(d) 102=2\cdot3\cdot17; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so \varphi(102)=102-70=32.

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