Question

The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29,2006 ) reported that in \(2006,20 \%\) of those in a random sample of 1112 Americans age 12 and older indicated that they owned an MP3 player. In a similar survey conducted in \(2005,\) only \(15 \%\) reported owning an \(\mathrm{MP} 3\) player. Suppose that the 2005 figure was also based on a random sample of size \(1112 .\) Estimate the difference in the proportion of Americans age 12 and older who owned an MP3 player in 2006 and the corresponding proportion for 2005 using a \(95 \%\) confidence interval. Is zero included in the interval? What does this tell you about the change in this proportion from 2005 to \(2006 ?\)

Short answer

The estimated difference in the proportion of Americans age 12 and older who owned an MP3 player between 2005 and 2006 ranges from 1.8% to 8.2%. As zero is not included in this interval, we can conclude that there was a statistically significant increase in the proportion of MP3 player owners from 2005 to 2006.

Step-by-step solution

Find the proportion for each year

In 2005, 15% of the 1112 surveyed individuals reported owning an MP3 player. This equates to \(0.15 * 1112 = 167\) individuals. In 2006, 20% of the 1112 individuals surveyed reported owning an MP3 player. This equates to \(0.2 * 1112 = 222\) individuals.

Calculate the standard error

The Standard error (SE) of a proportion is calculated by \(\sqrt {p * ( 1 - p ) / n}\), where p is the proportion and n is the sample size. For 2005, it's \(\sqrt(0.15 * (1 - 0.15) / 1112) = 0.011\). For 2006, it's \(\sqrt(0.2 * (1 - 0.2) / 1112) = 0.012\). The total standard error which is the difference is given by \(\sqrt{\(SE_1^{2}+SE_2^{2}}\) = 0.016.

Calculate the confidence interval

The difference in the proportions is \(0.2 - 0.15 = 0.05\). The 95% confidence interval around a proportion is calculated as 'proportion ± 1.96 * SE'. Therefore, the confidence interval for the change in proportions is \(0.05 \pm 1.96 * 0.016\), which gives a interval of (0.018, 0.082).

Analyze the interval

The interval does not contain zero, indicating a significant increase in the proportion of Americans age 12 and older who owned an MP3 player from 2005 to 2006. If zero were within the interval, it would suggest that the change in proportions isn't statistically significant, but that's not the case here.

Key concepts

Statistical Significance

When performing statistical analyses, it's crucial to distinguish between a genuine effect and a result that occurs by chance. This determination is where statistical significance comes into play.

Statistical significance is a term used to state that a certain statistic is reliable and not likely due to random variation. In the context of the MP3 player ownership, determining statistical significance helps to understand whether the observed difference in ownership percentages between 2005 and 2006 reflects a real change in the population of Americans age 12 and older, rather than random chance in sample selection.

The confidence interval calculated in the original problem helps us assess this. Since the 95% confidence interval for the difference in proportions does not include zero, it suggests the increase in ownership is indeed statistically significant. If this interval had contained zero, it would be possible that no true difference existed between the two years under consideration. Consequently, the lack of zero in the interval implies that we can be confident, within a 95% probability, that more individuals owned MP3 players in 2006 compared to 2005.

Standard Error

Understanding the concept of standard error (SE) is integral to interpreting statistical data accurately. SE gives us an idea of the accuracy of our estimate from a sample concerning the actual population parameter.

In the exercise, SE is computed using the formula for the standard error of the proportion, which is \(\sqrt {p * ( 1 - p ) / n}\), where \(p\) represents the sample proportion and \(n\) the sample size. The smaller the SE, the more precise is our estimate of the population proportion. In our case, the SE helps us measure the variability in the estimation of the proportion of Americans owning MP3 players for each year.

The total SE for the difference was calculated using the individual SEs from 2005 and 2006 and taking the square root of the sum of their squares. This calculation was part of determining the margin of error for the 95% confidence interval. The margin of error, when multiplied by a critical value (in this case, 1.96 for a 95% confidence level), gives us a range above and below the estimated difference that is likely to contain the true difference in the population proportions.

Proportion Estimation

Proportion estimation plays a vital role in statistics as it allows us to infer about a larger population based on sample data. In our exercise, the objective was to estimate the difference in proportions of Americans age 12 and older who owned an MP3 player in 2006 versus 2005.

From the samples for each year, we estimated that the proportion of MP3 player ownership rose from 15% to 20%. The absolute difference between these two sample proportions was 5% (or 0.05 when expressed as a decimal). However, since we are dealing with samples rather than the entire population, we cannot rely purely on this point estimate. Thus, we calculate a confidence interval, which provides a range of values within which we can be reasonably sure that the true difference in proportions lies.

To form this interval, we adjust the point estimate by adding and subtracting the margin of error, which incorporates the standard error and reflects the desired level of confidence. This process often uncovers insights about population parameters, which in this scenario pointed towards a meaningful increase in MP3 player ownership over the year-long span.