# Confidence intervals, again Several factors are involved in the creat

Confidence intervals, again Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true?
a) For a given sample size, reducing the margin of error will mean lower confidence.
b) For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample.
c) For a fixed margin of error, smaller samples will mean lower confidence.
d) For a given confidence level, a sample 9 times as large will make a margin of error one third as big.

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mauricio0815sh
Step 1
a. The confidence level increases as the margin of error increases and vice-versa.
Therefore, the statement "For a given sample size, reducing the margin of error will mean lower confidence' is true.
b. Margin of error:
The margin of error for the estimate of p is,
$$\displaystyle{M}{E}={z}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
The margin of error decreases as the sample size increases because the margin of error and the sample size are inversely proportional.
Hence, the variability would be less for larger sample which implies to smaller standard deviation that implies smaller margin of error.
Therefore, the statement "For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample" is true.
Step 2
c. For small sample, the variability would be more and the confidence interval becomes less successful in taking the true population proportion.
Therefore, the statement "For a fixed margin of error, smaller samples will mean lower confidence" is true.
d. The margin of error decreases as the sample size decreases because the margin of error and the sample size are inversely proportional.
Therefore, the statement "For a given confidence level, a sample 9 times as large will make a margin of error one third as big" is true.
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chumants6g

Step 1
(a) As the confidence level increases, we need to be more sure that the confidence interval contains the true population proportion and thus the confidence interval needs to contain more possible values, which means that the confidence interval be wider.
Since the margin of error is half the width of the confidence interval, the margin of error then needs to increase (not decrease) and thus the given statement is false.
Step 2
(b) As the sample larger, we have more information about the population proportion and thus our estimates will be more accurate.
The more accurate our estimates are, the $$n\Rightarrow er$$ the confidence interval.
Since the margin of error is half the width of the confidence interval, the margin of error then needs to decrease and thus the given statement is true.
Step 3
(c) As the sample becomes larges, we have more information about the population proportion and thus our estimates will be more accurate.
The more accurate our estimates are, the more confident we are that the population proportion has a certain margin of error and thus the confidence level increases, which means that the statement is true.
Step 4
(d) Formula for the margin of error:
$$\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
When the margin of error is halved (divided by 2):
$$\displaystyle{\frac{{{E}}}{{{2}}}}={z}_{{\frac{\alpha}{{2}}}}\cdot{\frac{{{1}}}{{{2}}}}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}={z}_{{\frac{\alpha}{{2}}}}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{2}^{{{2}}}{n}}}}}}={z}_{{\frac{\alpha}{{2}}}}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{4}{n}}}}}}$$
We then note that n is replaced by 4n and thus the sample size quadruples (not doubles) as the margin of error halves and thus the given statement is false.

nick1337

Step-by-step explanation:
Confidence interval is one which has centre as mean.  The width of the interval would be 2 times margin of error.
$$\text{Margin of error} = \text{Critical value} \cdot \text{Std error/sq rt of sample}$$
Hence if margin of error is lower for same n, then it means lower confidence level. Option a is right
Similarly, if bigger the sample size, lower is margin of error. Hence option b is right
c) False.  Smaller samples make margin of error big and hence confidence intervals bigger.
d) If n becomes 9n, then margin of error becomes $$\frac{1}{3}$$ Hence option d is right.