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Sever21 [200]
3 years ago
10

How large a sample n would you need to estimate p with margin of error 0.04 with 95% confidence? Assume that you don’t know anyt

hing about the value of p . 1037 256 601 423
Mathematics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

Since we don't know about p we can assume \hat p =0.5. And replacing into equation (b) the values from part a we got:  

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25  

And rounded up we have that n=601

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\hat p estimation for the sample proportion

n sample size selected

Confidence =0.95 or 95%

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Solution to the problem

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:  

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

Since we don't know about p we can assume \hat p =0.5. And replacing into equation (b) the values from part a we got:  

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25  

And rounded up we have that n=601

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a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

Z = \frac{X - \mu}{\sigma}

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X = 3.6

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.6 - 3.6}{0.4}

Z = 0

Z = 0 has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

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This is 1 subtracted by the pvalue of Z when X = 4.2. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

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0.0668 = 6.68% of the calls last more than 4.2 minutes

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This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

Z = \frac{X - \mu}{\sigma}

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This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

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Z = 3.5

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X = 3

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Z = -1.5

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At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

Z = \frac{X - \mu}{\sigma}

1.75 = \frac{X - 3.6}{0.4}

X - 3.6 = 0.4*1.75

X = 4.3

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