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LLS · Answer 1 · 2021-06-10T13:37:27+0000

Answer:

a. The smaller the sample size, the smaller the margin of error.

False if we see the value for the sample size is on the denominator and if we decrease the value of n then increase the value for the margin of error since the margin of error and the sample size are inversely proportional, with the estimated proportion and the confidence level fixed values.

b. If the margin of error is 0.05 and the observed proportion of red chips is 0.45, then the true population proportion is likely to be between 0.40 and 0.50.

Correct. The confidence interval for the population proportion is given by:

$\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}$

Or equivalently:

$\hat p \pm ME$

And if we replace on this case we got:

0.45-0.05=0.40

0.45 +0.05=0.50

So then the true proportion is between 0.4 and 0.5 for this case.

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

The population proportion have the following distribution

$p \sim N(p,\sqrt{(p(1-p))/(n)})$

The margin of error for the proportion interval is given by this formula:

$ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}$ (a)

Now if we analyze the statements one by one we have this: