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3 years ago

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Tom could grow potatoes, cabbages, and cauliflower on his farm. However, he has decided to grow only potatoes. What is Tom doing

?

2 answers:

enyata [817]3 years ago

8 0

Evaluating I think. rate my answer the brainliest pls.

dimaraw [331]3 years ago

3 0

He is growing potatoes, probably to sell them

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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