Question

In a study of 1,710 schoolchildren in Australia (Herald Sun, October 27,1994 ), 1,060 children indicated that they normally watch TV before school in the morning. (Interestingly, only \(35 \%\) of the parents said their children watched TV before school.) Construct and interpret a \(95 \%\) confidence interval for the proportion of all Australian children who say they watch TV before school. In order for the method used to construct the interval to be valid, what assumption about the sample must be reasonable?

Short answer

The \(95\%\) confidence interval for the proportion of all Australian children who say they watch TV before school is approximately \(0.596\) to \(0.643\) or \(59.6\%\) to \(64.3\%\). For the confidence interval method to be valid, it's necessary that the sample is random, independent, and sufficiently large (which it appears to be in this case).

Step-by-step solution

Calculate the Sample Proportion

To start, we need to calculate the sample proportion. The sample proportion (\(p\)) is the number of children who stated they watch TV before school divided by the total number of children. This is worked out as \(p = 1060 / 1710 = 0.62\)

Calculate the Standard Error

Next, we need to calculate the standard error (\(SE\)). The formula for standard error when dealing with proportions is \(\sqrt{p(1-p)/n}\), where \(n\) is the sample size. Substituting \(p = 0.62\) and \(n = 1710\), we get \(SE = \sqrt{0.62*(1-0.62)/1710} = 0.012\)

Establish the Confidence Level

The confidence level stated in the question is \(95\%\), which equates to a z-score of approximately 1.96 (based on a standard normal distribution table).

Construct the Confidence Interval

Now we have everything needed to form the \(95\%\) confidence interval. The confidence interval can be constructed as \(p ± Z*SE\). The lower bound (LB) and upper bound (UB) of the confidence interval can be calculated as follows: LB = \(p - Z*SE = 0.62 - 1.96*0.012 = 0.596\) and UB = \(p + Z*SE = 0.62 + 1.96*0.012 = 0.643\)

Interpret the Confidence Interval

The confidence interval ranges from \(0.596\) to \(0.643\). It means that the estimator is \(95\%\) confident that the true proportion of all Australian children who say they watch TV before school lies between \(59.6\%\) and \(64.3\%.\)

Evaluate the Assumption

For the confidence interval method to be valid, we have to assume that the sample drawn is random, each observation is independent of the others, and a certain condition known as the normal condition which says the sample size should be large enough must be reasonable. Given that the sample size is quite large (\(n = 1710\)), we can say these assumptions are reasonable under normal circumstances.

Key concepts

Sample Proportion

In statistics, the sample proportion is a vital concept that represents the fraction of individuals in a sample that have a particular characteristic. For example, in the scenario given, 1,710 schoolchildren were surveyed, and out of them, 1,060 children indicated that they watch TV before school. To calculate the sample proportion (\( p \)), you divide the number of children with the characteristic by the total sample size: \[ p = \frac{1060}{1710} = 0.62 \] This value is an estimate of the true proportion of the population that would answer similarly if every child were surveyed.

A key aspect of understanding the sample proportion is realizing it is an estimate derived from the sample data, and it will vary if different samples are taken. This variability is why constructing confidence intervals around the sample proportion can be crucial to estimating the true population proportion.

Standard Error

The standard error (SE) measures how far the sample proportion could be from the true population proportion. Essentially, it's an estimate of the standard deviation of the sampling distribution of the sample proportion. For proportions, the standard error is calculated using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the sample proportion and \( n \) is the sample size. In our example, substituting the values \( p = 0.62 \) and \( n = 1710 \) into the formula provides us with: \[ SE = \sqrt{0.62*(1-0.62)/1710} = 0.012 \] This SE helps us understand the variability in the estimate of the proportion and is crucial for constructing confidence intervals.

Understanding SE is fundamental because it reflects the precision of the sample proportion: a smaller SE suggests that the sample proportion is a more precise estimate of the true population proportion.

Confidence Level

The confidence level is an expression of how certain we are about the range in which the true population parameter lies. It's selected by the researcher and denotes the percentage of all possible samples that can be expected to include the population parameter. For instance, a 95% confidence level means that if we were to take 100 different samples and compute 100 different confidence intervals, we would expect about 95 of them to contain the true population proportion.

The confidence level correlates to a Z-score in the normal distribution, which is used to calculate the margin of error for the confidence interval. In the given study with a 95% confidence level, a Z-score of approximately 1.96 is used. This score is critical to finding the range in which we are 95% confident the true proportion will fall, and it reflects the degree of certainty we have in our interval estimate.

Normal Distribution

A normal distribution, often called the bell curve, is a continuous probability distribution that is symmetrical and has its mean, mode, and median all at the peak of the curve. The normal distribution plays a paramount role in statistics, especially concerning the construction of confidence intervals. It is underpinned by the Central Limit Theorem which states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the original population distribution.

When constructing a confidence interval for a sample proportion, we often use the standard normal distribution, which is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. This is applied in the calculation of Z-scores and ultimately, the confidence interval. The assumption that the sample proportion follows a normal distribution is crucial for the confidence interval to be valid.

Statistical Assumptions

In creating confidence intervals, certain statistical assumptions need to be met to ensure the results are valid. These include the assumption that the sample is a simple random sample from the population, implying that every individual has an equal chance of being chosen. Additionally, observations need to be independent; that is, the response from one individual should not influence the response from another.

Moreover, especially when dealing with proportions, the sample size should be large enough to justify the use of the normal distribution to model the sampling distribution (often this involves having at least 10 successes and 10 failures in your sample). This is known as the 'normal condition'. In the case of our example, with a sample size of 1,710, the assumption concerning the normalization condition is met, suggesting that the large sample size justifies the approximation of the distribution of the sample proportion to the normal distribution, making the construction of the confidence interval appropriate.