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3 years ago

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Nathaniel had a bag containing 75 marbles. He drew out a marble without looking and recorded the color. He then put the marble b

ack in the bag and drew another marble without looking and recorded its color. He continued this for 15 draws. The table shows his results.
what is the best estimate for the number of red marbles in the bag?

1 answer:

vlada-n [284]3 years ago

7 0

Answer:

Nathaniel had a bag containing 75 marbles. He drew out a marble without looking and recorded the color. He then put the marble back in the bag and drew another marble without looking and recorded its color. He continued this for 15 draws. The table shows his results.

what is the best estimate for the number of red marbles in the bag?

your question is incomplete

there is no table

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Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years.

Arada [10]

Answer:

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

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<u>Given</u>:

The satang coin from Thailand has a radius of about 7 mm.

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The diameter of the coin can be determined by multiply the radius by 2.

Thus, we have;

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$d=2(7)$

$d=14 \ mm$

Thus, the diameter of the coin is 14 mm.

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The circumference of the coin can be determined using the formula,

$C=\pi d$

Substituting $\pi=3.14$ and $d=14$ , we get;

$C=(3.14)(14)$

$C=43.96 \ mm$

Thus, the approximate value of the circumference of the coin is 43.96 mm

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