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Sauron [17]
3 years ago
8

Sam computed a 95% confidence interval for μ from a specific random sample. His confidence interval was 10.1 < μ < 12.2. H

e claims that the probability that μ is in this interval is 0.95. What is wrong with his claim?
A) Either μ is in the interval or it is not. Therefore, the probability that μ is in this interval is 0.95 or 0.05.
B) A probability can not be assigned to the event of μ falling in this interval.
C) Either μ is in the interval or it is not. Therefore, the probability that μ is in this interval is 0 or 1.
D) The probability that μ is in this interval is 0.05.
Mathematics
1 answer:
almond37 [142]3 years ago
3 0

Answer: C) Either μ is in the interval or it is not. Therefore, the probability that μ is in this interval is 0 or 1.

Step-by-step explanation: A 95% <u>Confidence</u> <u>interval</u> shows that there is a 95% confidence that the true parameter is between the lower and upper limits.

A CI is not a probability that the true parameter is in between the interval. The true parameter is either in the interval or not.

So, probability of falling between the limits is 0 (no chance of being in this interval) or 1 (100% possibility of being in this interval).

Then, "either μ is in the interval or it is not. therefore, the probability that μ is in this interval is 0 or 1." is correct.

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3 years ago
In a sample of nequals16 lichen​ specimens, the researchers found the mean and standard deviation of the amount of the radioacti
Dovator [93]

Answer:

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.001 microcurie per millilitre.

(c) The sample size necessary to obtain the desired estimate is 36.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive​ element, cesium-137 present in a sample of <em>n</em> = 16 lichen specimen are:

\bar x=0.009\\s=0.003

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.001 microcurie per millilitre of its true​ value, using a​ 95% confidence interval.

The (1 - <em>α</em>)% confidence interval for population mean (μ) is:

CI=\bar x \pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.001 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

Compute the sample size value as follows:

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\times s}{MOE}]^{2}

          =[\frac{1.96\times 0.003}{0.001}]^{2}

          =34.5744\\\approx36

Thus, the sample size necessary to obtain the desired estimate is 36.

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nlexa [21]

Answer:

Yes!

The solution to the equation is n = 2

Step-by-step explanation:

Given the equation

4\left(n\right)-3=-2\left(n\right)+9

remove the parentheses

4n-3=-2n+9

Add 3 to both sides

4n-3+3=-2n+9+3

Simplify

4n=-2n+12

Add 2n to both sides

4n+2n=-2n+12+2n

Simplify

6n=12

Divide both sides by 6

\frac{6n}{6}=\frac{12}{6}

Simplify

n=2

Therefore, the value of n = 2.

Hence,

Yes!

The solution to the equation is n = 2

5 0
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