Round 2,748,091 to the nearest 100,000

Mathematics

1 answer:

elixir [45]3 years ago

7 0

Answer:

I think it will be 2,700000

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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: