3. Last year, Lipton Tea Company conducted a mall intercept study at six regional malls around the country and found that 20% of the public preferred tea over coffee as a midafternoon hot drink. This year, Lipton wants to have a nationwide telephone survey performed with random digit dialing. What sample size should be used in this year’s study to achieve an accuracy level of ±2.5% at the 99% level of confidence? What about at the 95% level of confidence? (4 points)

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

Part a

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 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58[/tex]

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

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)

And on this case we have that [tex]ME =\pm 0.025[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.2(1-0.2)}{(\frac{0.025}{2.58})^2}=1704.04[/tex]

And rounded up we have that n=1705

Part 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 [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

[tex]n=\frac{0.2(1-0.2)}{(\frac{0.025}{1.96})^2}=983.45[/tex]

And rounded up we have that n=984