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3 years ago

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Students conducted a survey to see which cola was preferred most. People preferred cola A to cola B by a ratio of 9:7. If 108 pe

ople preferred cola A, how many people preferred cola B?
A:123
B:96
C:84
D:78

1 answer:

tigry1 [53]3 years ago

4 0

Answer:

c:84

Step-by-step explanation:

set your equasion to 9/7 = 108/x .solve the equasion and you get 84 for x

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Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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.

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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 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:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

Part a

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 $ME =\pm 0.02$ 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)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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