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The sample proportion is a fundamental concept when we're trying to estimate the true proportion of a population based on a sample. It is essentially a measure that represents how often a particular outcome happens within our sample group. In our case, 1060 out of 1710 children reported watching TV before school. To calculate the sample proportion, you divide the number of children who watch TV prior to school by the total number of children surveyed. Hence, the sample proportion here equates to

• Sample Proportion = Total TV watchers / Total children = 1060 / 1710 ≈ 0.620.

This proportion, approximately 0.620, is our estimate of the true proportion of all Australian schoolchildren who watch TV in the morning.

The standard error (SE) is crucial for understanding the variability of our sample estimate - the sample proportion - if the study were to be repeated multiple times. It gives us an idea of how much our sample proportion would vary from sample to sample.To calculate the standard error of the sample proportion, we use the formula:

• SE = \( \sqrt{\frac{p(1-p)}{n}} \)

where:

• \( p = 0.620 \)
• \( n = 1710 \)

Plugging these values into the formula, we get

• SE = \( \sqrt{\frac{0.620 \cdot 0.380}{1710}} \approx 0.012 \)

This tells us the standard deviation of our sample's proportion estimate, suggesting how far it will typically vary from the true proportion in the population if we were to conduct this study many times with different samples.

For statistical methods used to construct a confidence interval to be valid, certain assumptions need to be verified. These ensure the reliability and accuracy of the results. Here, the assumptions are:

• Simple Random Sample: Each child in the population should have an equal chance of being selected. This assumption guarantees that the sample reflects the broader population without bias.
• Large Sample Size: According to Bernoulli’s rule, the product of the sample size \( n \) and the sample proportion \( p \) should be greater than 5 (\( n \cdot p > 5 \)) and \( n \cdot (1-p) > 5 \). For our data: \( 1710 \cdot 0.620 > 5 \) and \( 1710 \cdot 0.380 > 5 \), both hold true, ensuring that the sample size is sufficiently large for the proportion to be approximately normally distributed.

These conditions help us establish a solid basis for estimating the wider population's behavior from our sample data.

Normal distribution is a key concept in statistics, especially related to confidence intervals. It describes how the values of a variable are expected to be distributed and is useful when estimating population parameters like the sample proportion. In this exercise, a normal distribution is assumed so that we can apply it to the confidence interval calculation.For a large enough sample size, the sampling distribution of the sample proportion can be approximated as normal. This approximation allows us to use the properties of the normal distribution to find the confidence interval. The concept here relies on the fact that:

• The mean of the normal distribution is the sample proportion \( p \), providing the "center" of our interval.
• The "spread" of the distribution, defined by the standard deviation, is derived from the standard error \( SE \).

Using these properties, we construct the 95% confidence interval by taking the sample proportion and adding and subtracting 1.96 times the standard error (SE) to capture the range that likely includes the true population proportion. This gives us a confidence interval that helps us statistically infer the true proportion of the entire population.