Question

A study by the Kaiser Family Foundation provided one of the first looks at media use among the very youngest children - those from 6 months to 6 years of age (Kaiser Family Foundation, \(2003,\) www.kff.org). The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." Because previous research indicated that children who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children ages 6 months to 3 years old. The second sample consisted of parents of children ages 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was 0.30 for the sample of children ages 6 months to 3 years and 0.43 for the sample of children ages 3 to 6 years old. Suppose that each of the two sample sizes was 100 . a. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of children ages 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of children ages 3 to 6 years who have a TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) large-sample confidence interval for the difference in the proportions for children ages 6 months to 3 years and for children ages 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

Short answer

The 95% confidence intervals for the proportions of children having a TV in their bedroom are [0.2098, 0.3902] for the age group 6 months to 3 years and [0.3339, 0.5261] for the age group 3 to 6 years. The overlapping of these intervals suggests that there might not be a significant difference between the two population proportions. The 95% confidence interval for the difference in these proportions is [-0.2624, 0.0024], which also suggests that there is no significant difference, but the proportion for the age group 3 to 6 years could be slightly higher.

Step-by-step solution

Construct confidence interval for 6 months to 3 years

For a single proportion, the confidence interval is calculated by the formula \(p \pm Z * \sqrt{p(1-p)/n}\). Here, p is the sample proportion which is 0.30, n is the sample size which is 100, and Z, the z-value is approximately 1.96 for a 95% confidence interval. Substituting these values, we get \(0.30 \pm 1.96 * \sqrt{0.30 * 0.70 / 100}\) which simplifies to \(0.30 \pm 0.0902\). This results in the interval \([0.2098, 0.3902]\).

Construct confidence interval for 3 to 6 years

Repeating the step 1 for the second sample with p as 0.43 we get \(0.43 \pm 1.96 * \sqrt{0.43 * 0.57 / 100}\) which simplifies to \(0.43 \pm0.0961\). This results in the interval \([0.3339, 0.5261]\).

Compare the intervals

The two intervals [0.2098, 0.3902] and [0.3339, 0.5261] do overlap. This suggests that there may be no significant difference between the two population proportions. However, we can't make this claim with certainty until we've constructed a confidence interval for the difference in proportions.

Construct the confidence interval for the difference in proportions

The confidence interval for the difference in proportions is calculated by the formula \((p1 - p2) \pm Z * \sqrt{p1(1 - p1)/n1 + p2(1 - p2)/n2}\). Substituting the respective values in the formula, we get \((0.30 - 0.43) \pm 1.96 * \sqrt{0.30 * 0.70 / 100 + 0.43 * 0.57 / 100}\), which simplifies to \(-0.13 \pm 0.1324\), resulting in the interval \([-0.2624, 0.0024]\). Since the interval includes 0, we can't say for certain if there is a significant difference between the proportions.

Compare the results

The comparison of the interval in Part (d) with the answer in Part (c) is consistent. In both the scenarios, we are unable to conclude with certainty whether there is a significant difference between the proportion of children having a TV in their bedroom in the two age groups. However, there is a slight indication that the proportion could be higher for the 3 to 6 years age group.

Key concepts

Large-Sample Confidence Interval

When working with a large sample size, as in the given exercise, statisticians use a large-sample confidence interval to estimate the true population parameter, such as a proportion. This type of confidence interval relies on the central limit theorem, which states that the distribution of sample means (or proportions) will be approximately normally distributed when the sample size is sufficiently large—typically a sample size greater than 30 is considered large.

The formula for constructing a large-sample confidence interval for a population proportion is:
\[ p \pm Z \times \sqrt{\frac{p(1-p)}{n}} \]
Here, \( p \) is the sample proportion, \( Z \) represents the z-score corresponding to the desired confidence level (which could be found in z-tables), and \( n \) is the sample size. This interval gives a range of values within which we can be confident (to a certain level, 95% in the exercise) that the true population proportion lies.

Understanding and correctly applying the large-sample confidence interval provides a foundation for interpreting the behaviors observed in a population.

Proportion Difference Confidence Interval

When we want to compare two different proportions, such as in this study on children's media use, we use a proportion difference confidence interval. This type of confidence interval allows us to estimate the difference between two population proportions based on sample data. It is particularly valuable when determining whether a significant difference exists between the two groups.

The formula used to calculate this confidence interval is:
\[ (p1 - p2) \pm Z \times \sqrt{\frac{p1(1 - p1)}{n1} + \frac{p2(1 - p2)}{n2}} \]
Here \( p1 \) and \( p2 \) are the sample proportions from the two groups, \( n1 \) and \( n2 \) are the sample sizes, and \( Z \) is the z-score for the chosen confidence level. The resulting interval provides insight into whether the observed difference in sample proportions is likely to reflect a true difference in the larger populations.

Sample Proportion

The sample proportion is a statistic that represents the fraction of individuals in a sample who have a certain characteristic. In this case, the characteristic is 'having a TV in their bedroom.' If we look at the given study, two sample proportions were found: 0.30 for children aged 6 months to 3 years, indicating 30% of the sample had TVs in their bedroom, and 0.43 for children aged 3 to 6 years, indicating 43% had TVs in their bedrooms.

To calculate a sample proportion, the formula is:
\[ p = \frac{X}{n} \]
where \( X \) is the number of individuals in the sample with the characteristic, and \( n \) is the total sample size. This simple yet essential statistic is critical for further analysis, such as creating confidence intervals or testing hypotheses about the population from which the sample was drawn.

Statistical Significance

The concept of statistical significance is central to hypothesis testing and simple to grasp. It tells us whether the results of a study or experiment reflect a true effect or if they are likely due to random chance. This is determined by a p-value, a probability score that shows the likelihood of obtaining our observed results, provided that the null hypothesis is true—it is no effect or no difference scenario.

If the p-value is lower than our predetermined significance level (commonly \( \alpha = 0.05 \)), we consider our results statistically significant and reject the null hypothesis. In the context of our confidence interval exercise, if a 95% confidence interval for the difference between two proportions does not include 0 (or some other value of interest, depending on the hypothesis), it is an indicator that there is likely a significant difference between the two population proportions. Conversely, as seen in the exercise where the interval does include 0, there is no statistical evidence to suggest the proportions from the two populations differ significantly.