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4 years ago

Help please!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

xeze [42]4 years ago

7 0

The answer is D.

The greater the pressure difference the faster the winds will move to attempt to equalize the pressures

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Step-by-step explanation:

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(hope that helped)

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A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their

Sever21 [200]

Answer:

A. Normal

B. Between 40.08 minutes and 43.92 minutes.

C. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, a normal distribution.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{12}{\sqrt{150}} = 1.92$

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

Between 40.08 minutes and 43.92 minutes.

Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

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3 years ago