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Chapter 9: Problem 58

In a study of 1710 schoolchildren in Australia (Herald Sun, October 27,1994 ), 1060 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 a \(95 \%\) confidence interval for the true proportion of Australian children who say they watch TV before school. What assumption about the sample must be true for the method used to construct the interval to be valid?

Short Answer

Expert verified
The \(95 \% \)confidence interval for the true proportion of Australian children who claim to watch TV before school is approximately between \(59.7 \% \) and \(64.3 \% \). This conclusion assumes that the sample is large enough that the sample proportion follows a normal distribution.

Step by step solution

01

Calculate the Sample Proportion (p虃)

The sample proportion is just the ratio of the number of positive outcomes (children who watch TV before school) to the total number of outcomes. In this case, it would be: \(p虃 = \frac{x}{n} = \frac{1060}{1710} = 0.620\). So, about \(62\%\) of the children watch television before school in the morning.
02

Calculating Z Value for Confidence Interval (z*)

For \(95 \% \) confidence interval, the Z value from the standard normal distribution is approximately \(1.96\). This value can be found using a standard normal distribution table or using a calculator that can calculate normal distribution percentiles.
03

Construct the Confidence Interval

The formula for the confidence interval of a proportion is given by the formula \(p虃 \pm z^{*} \sqrt{\frac{p虃(1-p虃)}{n}}\), where \(p虃\) is the estimated proportion, \(n\) is the sample size, and \(z^{*}\) is the Z-value for desired level of confidence. Plug in the values \(p虃=0.620, z^{*}=1.96, n=1710\) to get the interval as: \(0.620 \pm 1.96 * \sqrt{\frac{0.620 * 0.380}{1710}}\) = \(0.620 \pm 0.023\). So, it is expected that the true proportion of children in Australia who watch TV before school is in the range of \(59.7\% \)to \(64.3\%\).
04

The sample assumption validity

For the method used to construct the interval to be valid, there is an assumption that the sample proportion will follow a normal distribution. This is true for large sample sizes based on the Central Limit theorem. In this case, because the sample size (\(1710\)) is quite large, we can safely say that the normality assumption is valid.

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Key Concepts

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

Understanding Sample Proportion
When conducting studies or surveys, the sample proportion is a critical statistic. It represents the fraction of individuals in a sample that have a particular characteristic. For example, in a study involving 1710 children, if 1060 of them watch TV in the morning, the sample proportion watching TV would be the number of children watching TV divided by the total number surveyed, or approximately 0.620.

This figure not only gives a snapshot of the studied group's behavior but is also used to estimate the behavior of the larger population, in this case, to gauge the TV-watching habits of all Australian schoolchildren. To increase confidence in our estimates, one often calculates a confidence interval, which gives a range in which we believe the true population proportion lies.
The Role of Normal Distribution in Statistics
The normal distribution, often referred to as the bell curve, is a fundamental concept in statistics and refers to a distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. A perfect representation of this is the classic bell-shaped curve.
  • Many statistical methods are based on the assumption that the data follows a normal distribution due to its useful properties.
  • Knowing that our data follows a normal distribution allows us to predict the probability of events and create confident intervals for population parameters.
The normal distribution is crucial when we work with sample proportions because, under certain conditions, these proportions follow a pattern that closely resembles the normal distribution.
The Central Limit Theorem Explained
One of the cornerstones of statistics is the Central Limit Theorem. This theorem states that when independent random variables are added, their properly normalized sum tends toward a normal distribution (informally a bell curve) even if the original variables themselves are not normally distributed.

In practice, this means that the sample means or sample proportions will approximate a normal distribution as the sample size becomes larger, regardless of the population's distribution. The study example uses an adequately large sample size (1710 children); thus, we can safely say that the distribution of the sample proportion of children watching TV before school is roughly normal. This normality is critical in confidence interval calculations as it justifies using z-values derived from the standard normal distribution.
Interpreting Z-values in Statistics
A Z-value is a measure that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. In the context of confidence intervals, Z-values determine how many standard deviations away from the mean a certain percentile lies.
  • The Z-value for a 95% confidence interval is about 1.96. This means that if the data is normally distributed, approximately 95% of values should fall within 1.96 standard deviations of the mean, on either side.
  • When calculating confidence intervals for sample proportions, the formula incorporates the Z-value, the sample proportion, and the sample size, letting us make predictions about a population with a certain level of confidence.
Z-values thus connect our sample data with the broader context of the confidence interval, enabling statisticians to make inferences about the population as a whole.

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Most popular questions from this chapter

The two intervals (114.4,115.6) and \((114.1,\) 115.9 ) are confidence intervals (computed using the same sample data) for \(\mu=\) true average resonance frequency (in hertz) for all tennis rackets of a certain type. a. What is the value of the sample mean resonance frequency? b. The confidence level for one of these intervals is \(90 \%\) and for the other it is \(99 \%\). Which is which, and how can you tell?

If a hurricane was headed your way, would you evacuate? The headline of a press release issued lanuary 21, 2009 by the survey research company International Communications Research (icrsurvey,com )states, "Thirtyone Percent of People on High-Risk Coast Will Refuse Evacuation Order, Survey of Hurricane Preparedness Finds." This headline was based on a survey of 5046 adults who live within 20 miles of the coast in high hurricane risk counties of eight southern stares. In selecting the sample, care was taken to ensure that the sample would be representative of the population of coastal residents in these states. Use this information to estimate the proportion of coastal residents who would evacuate using a \(98 \%\) confidence interval. Write a few sentences interpreting the interval and the confidence level associated with the interval.

The article "The Assodation Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age" (Pediatrics 120051 : \(851-856\) ) reported the accompanying \(95 \%\) confidence intervals for average TV viewing time (in hours per day) for three different age groups. Confidence Interval Age Group \(=\) \(95 \%\) Less than (0.8,1.0) 12 12 to 23 2 months \begin{tabular}{l} (0.8,1.0) \\ (1.4,1.8) \\ (2.1,2.5) \\ \hline \end{tabular} \begin{tabular}{r} 12 to 2 \\ 24 to 3 \\ \hline \end{tabular} n 12 montns 5 months months 35 a. Suppose that the sample sizes for each of the three age group samples were equal. Based on the given confidence intervals, which of the age group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age group samples were equal, but that the three sample sizes might have been different. Which of the three age-group samples had the largest sample size? Explain your choice. c. The interval (.768,1.032) is either a \(90 \%\) confidence interval or a \(99 \%\) confidence interval for the mean TV viewing time computed using the sample data for children less than 12 months old. Is the confidence level for this interval \(90 \%\) or \(99 \%\) ? Explain your choice.

One thousand randomly selected adult Americans participated in a survey conducted by the Assodated Press (June 2006 ). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes okay. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. c. Comment on the apparent inconsistency in the responses given by the individuals in this sample.

The authors of the paper "Driven to Distraction" (Psychological Science 12001\(]=462-466\) ) describe an experiment to evaluate the effect of using a cell phone on reaction time, Subjects were asked to perform a simulated driving task while talking on a cell phone. While performing this task, occasional red and green lights flashed on the computer screen. If a green light flashed, subjects were to continue driving, but if a red light flashed, subjects were to brake as quickly as possible and the reaction time (in msec) was recorded. The following summary statistics are based on a graph that appeared in the paper: \(n=48 \quad \bar{x}=530 \quad s=70\) a. Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the mean time to react to a red light while talking on a cell phone. What assumption must be made in order to generalize this confidence interval to the population of all drivers? b. Suppose that the researchers wanted to estimate the mean reaction time to within 5 msec with \(95 \%\) confidence. Using the sample standard deviation from the study described as a preliminary estimate of the standard deviation of reaction times, compute the
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