Question

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 80% confident that his estimate is within six percentage points of the true population​percentage?
Complete parts​ (a) through​ (c) below.

1) Assume that nothing is known about the percentage of adults who have heard of the brand.
a.n=_________​(Round up to the nearest​ integer.)

2) Assume that a recent survey suggests that about 85​% of adults have heard of the brand.

b.n=_________(Round up to the nearest​ integer.)

​3) Given that the required sample size is relatively​ small, could he simply survey the adults at the nearest​ college?

Answer 1

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

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

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

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 80% of confidence, our significance level would be given by $\alpha=1-0.80=0.2$ and $\alpha/2 =0.1$. And the critical value would be given by:

$z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28$

Part 1

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 $ME =\pm 0.06$ or 6% points, 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 a prior estimate of $\het p$ we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77$  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is $\hat p =0.85$ so replacing the values into equation (b) we got:

$n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027$

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.