Question

The article "Kids Digital Day: Almost 8 Hours" (USA Today, January 20,2010 ) summarized a national survey of 2,002 Americans ages 8 to 18 . The sample was selected to be representative of Americans in this age group. a. Of those surveyed, 1,321 reported owning a cell phone. Use this information to construct and interpret a \(90 \%\) confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone. b. Of those surveyed, 1,522 reported owning an MP3 music player. Use this information to construct and interpret a \(90 \%\) confidence interval for the proportion of all Americans ages 8 to 18 who own an MP3 music player. c. Explain why the confidence interval from Part (b) is narrower than the confidence interval from Part (a) even though the confidence levels and the sample sizes used to calculate the two intervals were the same.

Short answer

The 90% confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone is about (0.638, 0.682) while the 90% confidence interval for who own an MP3 player is about (0.741, 0.779). The interval for owning a cell phone is wider since the sample proportion is closer to 0.5, resulting in a larger standard error.

Step-by-step solution

Calculate the proportion and standard errors

Firstly, calculate sample proportions and the standard errors for owning a cell phone and for owning an MP3 player. For the cell phone, the sample proportion \(p\) is 1321 divided by 2002, or about 0.66. The standard error is then calculated by taking \(\sqrt{p(1-p)/n} = \sqrt{(0.66)(0.34)/2002} \approx 0.011\). For MP3 player, the sample proportion \(p\) is 1522 divided by 2002, or about 0.760. The standard error, too, is determined as \(\sqrt{(0.760)(0.240)/2002} \approx 0.009\).

Construct the confidence intervals

Now, apply the formula for the confidence interval, which is \(p ± z*SE\), where \(z\) is the z-value from standard normal distribution associated with the desired level of confidence. For a 90% confidence interval, \(z = 1.645\). For both owning a phone and an MP3 player, calculate these intervals: For owning a cell phone, the interval is \(0.66 ± 1.645*0.011\), or about (0.638, 0.682). For owning an MP3 player, the interval is \(0.760 ± 1.645*0.009\), or about (0.741, 0.779).

Interpret the confidence intervals

Interpret the intervals in the context of the problem. The first interval suggests that we can be 90% confident that the true proportion of all Americans ages 8 to 18 who own a cell phone is between 0.638 and 0.682. The second interval indicates that we are 90% certain that the real proportion of all Americans of same ages who own an MP3 player is between 0.741 and 0.779.

Explain the difference in widths of the intervals

The width of a confidence interval depends on the standard error. If the sample proportion is close to 0.5, then the standard error and consequently the width of the interval is larger. This is the case with owning cell phone (the sample proportion was roughly 0.66). However, if the sample proportion is further from 0.5, like in the case of owning an MP3 player (the sample proportion was around 0.76), the standard error and the interval width becomes smaller.