Question

Suppose Brianna decided that the error margin of her 95% confidence interval was too large and wanted an error margin of .05 while maintaining a 95% confidence level. She should take another random sample of size n = _________. Use her estimate of π from problem 1.

Answer 1

Answer:

$n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296$ (b)  

Rounded up would be n =296

Step-by-step explanation:

Assuming this previous info:  She randomly interested in the p selects 100 engineering majors, and 26 of them said they earned an A in Calc 1.

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  

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

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)  

If solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)  

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:  

$z_{\alpha/2}=\pm 1.96$  

Thes estimated proportion is $\hat p = \frac{26}{100}=0.26$

And replacing we got:

$n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296$ (b)