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stealth61 [152]
3 years ago
6

Which pile has more pennies

Mathematics
1 answer:
Sloan [31]3 years ago
8 0
The left pile has more because if you the 6 is added there and pluses all the 12 makes it a greater number
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What is the mean of this data set? If necessary, round your answer to the
Marizza181 [45]

Answer:

40.1

Step-by-step explanation:

4 0
3 years ago
You wish to compute the 99% confidence interval for the population proportion. How large a sample should you draw to ensure that
Nataly_w [17]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47  

And rounded up we have that n=99

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

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 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

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 0.13 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)  

We assume the value for \hat p =0.5 since we don't have previous info. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47  

And rounded up we have that n=99

4 0
3 years ago
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In the past, the mean running time for a certain type of flashlight battery has been 9.7 hours. The manufacturer has introduced
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Answer:

Step-by-step explanation:

From the information given:

\mathbf{H_o: \mu = 9.7 \ hours}

\mathbf{H_1: \mu > 9.7 \ hours}

The type 1 error is rejecting \mathbf{H_o} when

The meaning of Type 1 error is rejecting the claim that the mean running time is 9.7 hours when actually the mean running time is greater than 9. 7 hour.

6 0
2 years ago
....................
nadezda [96]

Answer:

.

Step-by-step explanation:

4 0
1 year ago
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3 stickers in each bag
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