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zysi [14]
3 years ago
7

The distribution of the amount of money in savings accounts for Florida State students has an average of 1,200 dollars and a sta

ndard deviation of 900 dollars. Suppose that we take a random sample of 10 Florida State students and ask them how much they have in their savings account. The sampling distribution of the sample mean amount of money in a savings account is
Mathematics
1 answer:
Anestetic [448]3 years ago
7 0

Answer:

By the Central Limit Theorem, the sampling distribution of the sample mean amount of money in a savings account is approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes 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.

Average of 1,200 dollars and a standard deviation of 900 dollars.

This means that \mu = 1200, \sigma = 900

Sample of 10.

This means that n = 10, s = \frac{900}{\sqrt{10}} = 284.6

The sampling distribution of the sample mean amount of money in a savings account is

By the Central Limit Theorem, approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

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3 years ago
A firm has a debt to asset ratio of 75%, $264,000 in debt, and net income of $49,280. calculate return on equity
svlad2 [7]
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2 years ago
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3 years ago
Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

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Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

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t(s) = 2,3237

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t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
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Answer:

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2 years ago
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