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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 ?

Step by step solution

01

Determine the sample proportions

First we need to determine the proportions for both years. In 2006, 20% of 1112 people reportedly owned an MP3 player, so the sample proportion is 0.20. Similarly, in 2005, the sample proportion is 0.15. So we have \( p_1 = 0.20 \) and \( p_2 = 0.15 \). Of course, we also know the total number of people surveyed, \( n_1 = n_2 = 1112 \).

02

Calculate the difference and standard error

We are interested in the difference \(d = p_1 - p_2\), so we calculate that and find \(d = 0.20 - 0.15 = 0.05\). Next, we need to calculate the standard error (SE) of the difference. The formula for this is \(\sqrt{{p_1(1 - p_1) / n_1 + p_2(1 - p_2) / n_2}}\), resulting in an approximate SE of 0.019.

03

Construct the confidence interval

A \(95\%\) confidence interval is defined as the sample difference \(±\) the margin of error (ME). The ME can be calculated by multiplying the standard error by the z-score corresponding to the desired level of confidence. For a \(95\%\) confidence level, the z-score is approximately 1.96. Therefore, the ME = 1.96 * SE = 1.96 * 0.019 = 0.037. The confidence interval then becomes 0.05 \(±\) 0.037, or approximately \([0.013, 0.087]\).

04

Interpret the results

The resulting confidence interval doesn't include zero, which suggests that the increase in MP3 player ownership from 2005 to 2006 is statistically significant. In other words, the proportion of Americans age 12 and older who owned MP3 players likely did increase during this one-year period.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion

A sample proportion is a metric used to describe the proportion of a characteristic within a subset of your total population. This is useful in surveys or research where measuring every individual in a population is impractical. For example, in a survey conducted in 2006, 20% of the sample reported owning an MP3 player. Here, the sample proportion is represented as 0.20. Similarly, the sample proportion from 2005 is 0.15, indicating 15% ownership. To calculate a sample proportion, follow these steps:

• Identify the characteristic of interest, such as MP3 ownership.
• Divide the number of individuals with the characteristic by the total sample size. For example, if 222 people out of 1112 owned an MP3 player, divide 222 by 1112 to get 0.20.
• Use this proportion in further statistical analysis to generalize to the broader population.

Standard Error

The standard error (SE) tells us how much variability we can expect in the sample proportion estimates if we repeated the survey many times. It acts as a measure of precision or spread of the sample proportion.To calculate the standard error of difference between two proportions, use this formula:\[ SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} \]Where:

• \( p_1 \) and \( p_2 \) are the sample proportions from 2006 and 2005, respectively.
• \( n_1 \) and \( n_2 \) are the sample sizes for each year, in this case, both are 1112.

Using the given proportions, the calculated SE provides an estimate of 0.019. The smaller the standard error, the more precise the estimate of the sample proportion.

Statistical Significance

Statistical significance helps us determine if the observed difference or relationship in the data could have occurred by random chance. In the context of proportions, it tells us whether the change in ownership from 2005 to 2006 is likely to reflect a true shift in the population.To check for statistical significance:

• Look at the confidence interval, which is constructed using the standard error and the z-score for the chosen confidence level. A 95% confidence interval means you are 95% confident that the interval contains the true difference.
• If the confidence interval excludes zero, it suggests there is a statistically significant difference. In this exercise, the interval is \([0.013, 0.087]\), meaning zero is not included. This suggests a statistically significant increase in MP3 player ownership from 2005 to 2006.

Proportion Difference Estimation

Proportion difference estimation is used to determine the difference between two proportions from different groups or time periods. It provides insights into changes or differences in a particular characteristic, such as MP3 player ownership, over time or between groups. Here's how to estimate the difference between two proportions:

• Calculate the difference between the two sample proportions. For instance, in this exercise, it’s the ownership difference from 2006 to 2005, or 0.20 - 0.15, resulting in 0.05.
• Construct a confidence interval around this difference to understand the range within which the true difference may lie. The standard error and the confidence level contribute to computing the margin of error used in forming the interval.
• The confidence interval \([0.013, 0.087]\) indicates that the true proportion difference likely lies within this range, with the center at 0.05. This helped determine that there was a reliable increase in ownership between these years.