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Eduardwww [97]
3 years ago
15

The safety inspector in a large city wants to estimate the proportion of buildings in the city that are in violation of fire cod

es. A random sample of 40 buildings is chosen for inspection, and 4 of them are found to have fire code violations. Estimate the proportion of buildings in the city that have fire code violations, and find the uncertainty in the estimate.
Mathematics
1 answer:
Drupady [299]3 years ago
7 0

Answer:

Proportion of buildings in the city that have fire code violations: 10%

Uncertainty 0.3 = 30%

Step-by-step explanation:

This situation complies with a binomial model where p is the probability of finding a building having fire code violations and q=1-p

Since in random sample of 40 buildings chosen for inspection 4 of them are found to have fire code violations, the proportion  is 4/40 = 0.1 or 10% and  p=0.1 as well.

<em>In a</em> <em>binomial model the standard deviation s is: </em>

<em>\large s=\sqrt{npq} </em>

<em>where n is the sample size</em>. So

\large s=\sqrt{40*0.1*0.9}=1.8974

and the uncertainty is the standard error SE

\large SE=\frac{s}{\sqrt{n}}=\frac{1.8974}{\sqrt{40}}=0.3=30\%

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Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi
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Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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".  

Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

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

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

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This is never true. In a parallelogram, the diagonals will always bisect each other. Thus, each of these segments would always have to be equal. 
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