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Young Americans, Part. II. About \(25 \%\) of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed. (b) In order for the the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40 . (c) A random sample of 50 young Americans where \(20 \%\) have delayed starting a family due to the continued economic slump would be considered unusual. (d) A random sample of 150 young Americans where \(20 \%\) have delayed starting a family due to the continued economic shmp would be considered unusual. (e) Tripling the sample size will reduce the standard error of the sample proportion by one-third.

Short Answer

Expert verified
(a) True, (b) True, (c) False, (d) False, (e) False.

Step by step solution

01

Analyze Skewness with Sample Size 12

The skewness of the sample proportions depends on both the proportion \( p = 0.25 \) of the population and the sample size \( n = 12 \). A sample proportion is approximately normally distributed if both \( np \) and \( n(1-p) \) are greater than 5. For \( n = 12 \), \( np = 12 \times 0.25 = 3 \) and \( n(1-p) = 12 \times 0.75 = 9 \). Since \( np < 5 \), it indicates a right-skewed distribution.
02

Determine when the Distribution is Approximately Normal

The Central Limit Theorem states that the sampling distribution of the sample proportion is approximately normal if the sample size \( n \) is large enough such that both \( np \) and \( n(1-p) \) are greater than 5. Let's check with \( n = 40 \): \( np = 40 \times 0.25 = 10 \) and \( n(1-p) = 40 \times 0.75 = 30 \), both are greater than 5, so a sample size of 40 or greater is sufficient for normal approximation.
03

Consider Sample Proportion Unusualness (n=50)

The standard error (SE) of the sample proportion is calculated as \( SE = \sqrt{\frac{p(1-p)}{n}} \). For \( n = 50 \), \( SE = \sqrt{\frac{0.25 \times 0.75}{50}} \). Calculate \( SE \): \( SE = \sqrt{0.00375} = 0.0612 \). To determine if it's unusual, calculate the expected proportion range: mean \( p = 0.25 \), and 95% CI is \( 0.25 \pm 2\times0.0612 = (0.1276, 0.3724) \). 20% (0.20) is within this range, thus it's not unusual.
04

Consider Sample Proportion Unusualness (n=150)

For \( n = 150 \), \( SE = \sqrt{\frac{0.25 \times 0.75}{150}} = 0.035 \). Calculate the range again for 95% CI: mean \( p = 0.25 \), \( 0.25 \pm 2\times0.035 = (0.18, 0.32) \). 20% (0.20) is within this range, thus a sample proportion of 20% is not unusual.
05

Effect of Tripling the Sample Size on Standard Error

The standard error \( SE \) of the sample proportion \( \sqrt{\frac{p(1-p)}{n}} \) is inversely proportional to the square root of the sample size \( n \). If the sample size is tripled \( k \to 3k \), the standard error becomes \( \frac{1}{\sqrt{3}} \) of the original standard error. \( \sqrt{3} \approx 1.73 \), so the standard error is reduced by about a factor of 1/\( 1.73 \), which is not exactly one-third.

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Key Concepts

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

Sample Proportion
The concept of sample proportion is crucial when analyzing data collected from a sample rather than an entire population. If you want to understand the proportion, think of it as a fraction of the sample showing a particular trait or characteristic. For instance, in the case of young Americans delaying families, if we have a sample of 100 individuals, and 25 of them delayed starting a family, the sample proportion is 0.25.

In statistical studies, this proportion helps us to make inferences about the population as a whole. When citing sample proportions, it's essential to note whether the sample accurately reflects the broader population. This ensures the validity of any conclusions drawn. In calculations, the sample proportion is expressed as \( p = \frac{x}{n} \), where \( x \) is the number of successful outcomes and \( n \) is the sample size.

It's important to consider the Central Limit Theorem while analyzing sample proportions, as this theorem helps predict the behavior of the sample mean and proportions in large samples, guiding us in determining their normality.
Standard Error
Standard error (SE) is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation. Specifically for sample proportions, the standard error gives an indication of how much the proportion will vary from one sample to another. It is calculated using the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( p \) is the sample proportion, and \( n \) is the sample size.

A smaller SE means the sample proportion is more likely to be similar across different samples, making it a reliable indicator of the population proportion. This is why it's crucial to have an adequate sample size; a larger sample size will typically result in a smaller standard error, making your findings more precise.
  • Increasing the sample size decreases the standard error, thus providing a closer estimation of the population parameter.
  • If the variation within the sample reduces, it suggests that one can confidently generalize the sample findings for the whole population.
Skewness
Skewness measures asymmetry in a statistical distribution. In simpler terms, it indicates whether data points are stretched more on one side (left or right) of the mean, showing an unsymmetrical tail. For sample proportions, skewness might occur in distributions with relatively small sample sizes or when the proportion is near 0 or 1.

A right-skewed distribution means more data points are concentrated on the left, with a tail on the right, which leads to the mean being greater than the median. For a sample size of 12, if the product of \( np \) and \( n(1-p) \) is less than 5, the sampling distribution of the sample proportion tends to be right-skewed. This tells us the data spreads more on the lower side, indicating potential limited variability seen in small samples.
  • Right-skewness can indicate outliers or a long tail on the right side of the distribution.
  • Understanding skewness helps in interpreting how sample data might underestimate or overestimate population parameters.
Confidence Interval
A confidence interval provides a range that acts as an estimate of a population parameter, calculated from a given set of sample data. In statistics, confidence intervals give us an idea of how uncertain a sample statistic is, like the sample proportion, when estimating the population proportion.

This range is computed with a confidence level, which typically could be 95% or 99%, indicating that the true population parameter lies within the interval a certain percentage of the time. The confidence interval can be calculated as \( \hat{p} \pm Z \times SE \), where \( \hat{p} \) is the sample proportion and \( Z \) is the Z-score corresponding to the desired confidence level.
  • A 95% confidence interval means there is a 95% chance the population parameter will fall within this interval.
  • This range helps statisticians to estimate how variable the sample proportion might be, enabling accurately deduced insights about the population.

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Most popular questions from this chapter

Sleep deprived transportation workers. The National Sleep Foundation conducted a survey on the sleep habits of randomly sampled transportation workers and a control sample of non-transportation workers. The results of the survey are shown below. \begin{tabular}{lccccc} & \multicolumn{4}{c} { Transportation Professionals } \\ \cline { 3 - 6 } & & & Truck & Train & Bux/1axi/Limo \\ & Control & Pilots & Drivers & Operators & Drivers \\ \hline Less than 6 hours of sleep & 35 & 19 & 35 & 29 & 21 \\ 6 to 8 hours of sleep & 193 & 132 & 117 & 119 & 131 \\ More than 8 hours & 64 & 51 & 51 & 32 & 58 \\ \hline Total & 292 & 202 & 203 & 180 & 210 \end{tabular} Conduct a hypothesis test to evaluate if these data provide evidence of a difference between the proportions of truck drivers and non-transportation workers (the control group) who get less than 6 hours of sleep per day, i.e. are considered sleep deprived.

True or false, Part I. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) The chi-square distribution, just like the normal distribution, has two parameters, mean and standard deviation. (b) The chi-square distribution is always right skewed, regardless of the value of the degrees of freedom parameter. (c) The chi-square statistic is always positive. (d) As the degrees of freedom increases, the shape of the chi-square distribution becomes more skewed.

Offshore drilling, Part II. Results of a poll evaluating support for drilling for oil and natural gas off the coast of California were introduced in Exercise \(3.29 .\) \begin{tabular}{lcc} & \multicolumn{2}{c} { College Grad } \\ \cline { 2 - 3 } & Yes & No \\ \hline Support & 154 & 132 \\ Oppose & 180 & 126 \\ Do not know & 104 & 131 \\ \hline Total & 438 & 389 \end{tabular} (a) What percent of college graduates and what percent of the non-college graduates in this sample support drilling for oil and natural gas off the Coast of California? (b) Conduet a hypothesis test to determine if the data provide strong evidence that the proportion of college graduates who support off-shore drilling in California is different than that of noncollege graduates.

Life after college. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. Suppose we conduct a survey and find out that 348 of the 400 randomly sampled graduates found jobs. The graduating class under consideration included over 4500 students. (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions for constructing a confidence interval based on these data are met. (c) Calculate a \(95 \%\) confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data. (d) What does "95\% confidence" mean? (e) Now calculate a \(99 \%\) confidence interval for the same parameter and interpret it in the context of the data. (f) Compare the widths of the \(95 \%\) and \(99 \%\) confidence intervals. Which one is wider? Explain.

Public Option, Part. II1. Exercise 3.13 presents the results of a poll evaluating support for the health care public option plan in 2009. \(70 \%\) of 819 Democrats and \(42 \%\) of 783 Independents support the public option. (a) Calculate a \(95 \%\) confidence interval for the difference between \(\left(p_{D}-p_{1}\right)\) and interpret it in this context. We have already checked conditions for you. (b) True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it is more likely that the Democrat would support the public option than the Independent.

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