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melisa1 [442]
3 years ago
12

In Test A, suppose you make 100 experimental measurements of some quantity and then calculate mean, standard deviation, and stan

dard error of the numbers you obtain. In Test B, suppose you make 1000 experimental measurements of the same quantity and you again calculate mean, standard deviation, and standard error.
How would you expect the mean, standard deviation, and standard errors found in Test A and Test B to compare? (You can use the Excel simulation and change the "Number of Experiments" parameter to investigate this yourself.
A.) I would expect the means, the standard deviations, and the standard errors in Test A and Test B to be about the same.
B.) I would expect the means of the two tests to be about the same, but both the standard deviation and the standard error in Test B should be smaller than in Test A
C.) I would expect the means of the two tests to be about the same, but both the standard deviation and the standard error in Test B should be bigger than in Test A
D.) I would expect the means and standard deviations in the two test to be about the same, but the standard error in Test B should be smaller than in Test A.
E.) would expect the means and standard deviations in the two test to be about the same, but the standard error in Test B should be larger than in Test A.
Mathematics
1 answer:
Scrat [10]3 years ago
8 0

Answer:

D.) I would expect the means and standard deviations in the two test to be about the same, but the standard error in Test B should be smaller than in Test A.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

By the Central Limit Theorem:

Both tests, A and B, are of the same quantity, so they have the same mean and standard deviation. However, since the standard error is inversely proportional to the sample size, and test B has a greater sample size than test A, the standard error in test B should be lower, and thus, the correct answer is given by option D.

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Given:

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3 years ago
The athlete’s salary, in thousands, for the first two years is $400 and $400(1.05). Explain how to find her salary for each of t
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where 
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We can infer from our values, that the current term of the sequence is 400(1.5), so a_{n-1}=400(1.5). That leaves 400 as the previous term, so a_{n-1}=400. Lets replace those values in our formula to find r:
r= \frac{a_{n} }{a_{n-1}}
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Now that we have our common ratio, we can replace it in our formula for the nth term to find the athlete's salary for each of the next three years. Notice that the first term of our sequence is $400, so a_{1}=400
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a_{4}=400(1.05)^{4-1}
a_{4}=400(1.05)^{3}
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a_{5}=400(1.05)^{5-4}
a_{5}=400(1.05)^4
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We can conclude that the athlete's salary for each of the next three years is: $441,$463.05,486.2025 respectively. Also, those vales for a geometric sequence because they share a common ratio, (1.05).
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