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Elden [556K]
2 years ago
13

To compare the average amount of time that canadians and americans spend commuting, a researcher collects a sample of canadians

and americans. The canadians spend an average of hours a week commuting, with a standard deviation hours. The mean and standard deviation for the sample of americans is hours and hours, respectively. The standard error of the difference of sample means is:.
SAT
1 answer:
yKpoI14uk [10]2 years ago
5 0

The standard error of the difference of sample means is 0.444

From the complete question, we have the following parameters

<u>Canadians</u>

  • Sample size = 50
  • Mean = 4.6
  • Standard deviation = 2.9

<u>Americans</u>

  • Sample size = 60
  • Mean = 5.2
  • Standard deviation = 1.3

The standard error of a sample is the quotient of the standard deviation and the square root of the sample size.

This is represented as:

SE = \frac{\sigma}{\sqrt n}

The standard error of the Canadian sample is:

SE_1 = \frac{2.9}{\sqrt{50}}

So, we have:

SE_1 = 0.41

The standard error of the American sample is:

SE_2 = \frac{1.3}{\sqrt{60}}

So, we have:

SE_2 = 0.17

The standard error of the difference of sample means is then calculated as:

SE= \sqrt{SE_1^2 + SE_2^2}

This gives

SE= \sqrt{0.41^2 + 0.17^2}

SE= \sqrt{0.197}

Take square roots

SE= 0.444

Hence, the standard error of the difference of sample means is 0.444

Read more about standard errors at:

brainly.com/question/6851971

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The community health nurse is conducting a breast cancer screening clinic in a local neighborhood and is providing sessions on b
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1 year ago
If sin(1/x^2+1) is an antiderivative for f(x) then.
STatiana [176]

The limit from 1 to 2 of the given antiderivative is; -0.19865

<h3>What is the Limit of the Integral?</h3>

We are given the antiderivative of f(x) as sin(1/(x² + 1)). Thus, to find the limit from 1 to 2, we will solve as;

\int\limits^2_1 {(sin\frac{1}{x^{2} + 1} )} \, dx = {(sin\frac{1}{2^{2} + 1} )}  - {(sin\frac{1}{1^{2} + 1} )}

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⇒ 0.19866 - 0.47942

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Complete Question is;

If sin(1/(x² + 1)) is an anti derivative for f(x), then what is the limit of f(x)dx from 1 to 2?

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