Question

1. A 2002 poll reported that 61​% of people worried that they would be exposed to SARS.

Answer 1

Answer:

Part 1

a)$ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956$  

b) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502$

c) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243$

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

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".  

The population proportion have the following distribution

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

Part 1

For this case in order to find the margin of error we need to assume a confidence level fixed, let's assume 95% for example. The Margin of error is given by this formula:

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

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 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 $\hat p = 0.61$ and we are interested in order to find the value of ME.

So we can replace for each case and see what we got:

a)$ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956$  

b) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502$

c) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243$

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.