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AnnZ [28]
3 years ago
7

Suppose you are spinning pennies to test whether you get biased results. When you reject the null hypothesis when it is actually

true, that is often called the first kind of error. The second kind of error is when the null is false and you fail to reject. Report the first kind of error and the second kind of errorThe null hypothesis is that the penny is ........(not biased/biased). The first kind of error is saying the penny is ...,,(biased/not biased) when in fact the penny is .....(biased/not biased). The second kind of error is saying the pennyIs .....(not biased/biased) when in fact the penny is .....(biased/not biased).
Mathematics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

Step-by-step explanation: i'm not really sure ether

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Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years.
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a) 0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799

0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847

The 95% confidence interval would be given by (0.799;0.847)

b) n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

c) n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

Part a

\hat p=\frac{823}{1000}=0.823

In order to find the critical value 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 value would be given by:

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

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799

0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847

The 95% confidence interval would be given by (0.799;0.847)

Part b

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 0.03 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)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

Part c

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

5 0
3 years ago
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