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3 years ago

How to solve this problem

Mathematics

1 answer:

iragen [17]3 years ago

5 0

1.) 737.303-43
Subtract
Final Answer: 694.303

2.) 275.6-6.9
Subtract
Final Answer: 268.7

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a) $n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83$

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724$

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :

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

Part a

The estimated proportion for this case is $\hat p =0.76$

Our interval is at 90% of confidence, and the significance level is given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The critical values for this case are:

$z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64$

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)

The margin of error desired is given $ME =\pm 0.01$ 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)

Replacing we got:

$n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83$

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724$

And rounded up we have that n=6724

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Step-by-step explanation:

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