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Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is \(8.0 \%\), while this proportion is \(8.8 \%\) for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a \(95 \%\) confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. \(^{44}\)

Short Answer

Expert verified
The 95% confidence interval is [-0.0178, 0.0018].

Step by step solution

01

Define Sample Proportions and Sizes

Identify the key values: The sample proportion of Californians who are sleep deprived is \( p_1 = 0.08 \), and the sample size is \( n_1 = 11,545 \). For Oregonians, the sample proportion is \( p_2 = 0.088 \), with a sample size of \( n_2 = 4,691 \).
02

Calculate the Difference in Sample Proportions

The difference in sample proportions is calculated as \( p_1 - p_2 = 0.08 - 0.088 = -0.008 \).
03

Calculate Standard Error for Difference in Proportions

Calculate the standard error (SE) using the formula: \[ SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \] Substitute the values: \[ SE = \sqrt{\frac{0.08 \times 0.92}{11,545} + \frac{0.088 \times 0.912}{4,691}} \] Computing gives:\[ SE \approx 0.005 \].
04

Find Z-Score for Confidence Level

For a \(95\%\) confidence level, the z-score is approximately \(1.96\).
05

Calculate Confidence Interval

Use the formula:\[ CI = \text{Difference} \pm (Z \times SE) \] Substitute the known values:\[ CI = -0.008 \pm 1.96 \times 0.005 \] This yields:\[ CI = -0.008 \pm 0.0098 \] Thus, the confidence interval is \([-0.0178, 0.0018]\).
06

Interpret the Confidence Interval

The confidence interval \([-0.0178, 0.0018]\) suggests that we are \(95\%\) confident that the proportion of sleep-deprived Californians is between \(1.78\%\) less and \(0.18\%\) more than the proportion of sleep-deprived Oregonians.

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Key Concepts

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

Sleep Deprivation
Sleep deprivation occurs when individuals do not get the required amount of sleep. This can result from lifestyle choices, work demands, or health conditions. Sleep is vital for health and well-being, affecting brain function and overall mood. When large numbers of a population experience sleep deprivation, it becomes a public health issue.
In our exercise, the Centers for Disease Control and Prevention conducted a study to understand sleep deprivation among residents in California and Oregon. The findings showed that 8.0% of Californians and 8.8% of Oregonians reported insufficient sleep.
These reports are statistically significant as they help in understanding and addressing potential public health concerns regarding sleep in these states. Public health officials can use this data to formulate strategies to promote better sleep habits among the population.
Sample Proportions
The concept of sample proportions is crucial for estimating how a particular attribute or behavior is distributed across a population. A sample proportion is calculated by dividing the number of successful outcomes by the total number of observations in the sample.
For instance, in the sleep deprivation study, the sample proportion of sleep-deprived individuals in California was given by dividing the number of sleep-deprived residents by the total number of surveyed individuals in this state: \( p_1 = \frac{0.08 \times 11,545}{11,545} = 0.08 \).
Similarly, for Oregon, the sample proportion was \( p_2 = 0.088 \). It’s important to remember that these proportions are estimates of the larger population's sleep habits.
  • They provide a snapshot of focused data.
  • They're instrumental in analyzing differences between groups.
Standard Error
The standard error (SE) is a key statistical measure that estimates the variability or dispersion of sample proportions. It is especially important in determining the reliability of the sample mean's representation of the population mean.
In the context of sleep deprivation studies, SE helps to gauge how much the sample proportion might differ from the actual population proportion.
The formula used to calculate the standard error for our difference in proportions is:\[ SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \] Substituting the given values results in: \( SE \approx 0.005 \).
  • SE gives insight into data variability.
  • It is a foundational element for constructing confidence intervals.
Z-Score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations from the mean, providing a way to compare results from different tests or indicators. The z-score shows how data point compares to the normal distribution.
For a 95% confidence level, a typical z-score is 1.96, which corresponds to the middle 95% of a standard normal distribution.
In our sleep deprivation example, the z-score was used to calculate the confidence interval, showing the difference in sleep deprivation proportions between California and Oregon:\( CI = \text{Difference} \pm (Z \times SE) \).Using: \( Z = 1.96 \times SE = 0.005 \), gives the confidence interval: \([-0.0178, 0.0018]\).
  • Z-score helps in standardizing data results across datasets.
  • It's essential for determining confidence intervals in statistical analysis.

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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.

Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use and autism surveyed the mothers of a random sample of children aged \(24-60\) months with autism and conducted another separate random sample for children with typical development. The table below shows the number of mothers in each group who did and did not use prenatal vitamins during the three months before pregnancy (periconceptional period). 40 \begin{tabular}{llccc} & \multicolumn{4}{c} { Autism } \\ \cline { 3 - 4 } & & Autism Typical development & Total \\ \cline { 2 - 5 } Periconceptional & No vitamin & 111 & 70 & 181 \\ prenatal vitamin & Vitamin & 143 & 159 & 302 \\ \cline { 2 - 5 } & Total & 254 & 229 & 483 \end{tabular} (a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three months before pregnancy and autism. (b) Complete the hypothesis test and state an appropriate conclusion. (Reminder: verify any necessary conditions for the test.) (c) A New York Times article reporting on this study was titled "Prenatal Vitamins May Ward Off Autism", Do you find the title of this article to be appropriate? Explain your answer. Additionally, propose an alternative title.

Life rating in Greece. Greece has faced a severe economic crisis since the end of \(2009 .\) A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that \(25 \%\) of them said they would rate their lives poorly enough to be considered "suffering". (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions required for constructing a confidence interval based on these data are met. (c) Construct a \(95 \%\) confidence interval for the proportion of Greeks who are "suffering". (d) Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level. (e) Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample.

Open source textbook. A professor using an open source introductory statistics book predicts that \(60 \%\) of the students will purchase a hard copy of the book, \(25 \%\) will print it out from the web, and \(15 \%\) will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online. (a) State the hypotheses for testing if the professor's predictions were inaccurate. (b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verify they are satisfied. (d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context.

Gender and color preference. A 2001 study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A \(95 \%\) confidence interval for the difference between the proportions of males and females whose favorite color is black ( \(p_{\text {male }}-p\) female \()\) was calculated to be (0.02,0.06) . Based on this information, determine if the following statements are true or false, and explain your reasoning for each statement you identify as false. 2 (a) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) lower to \(6 \%\) higher than the true proportion of females whose favorite color is black. (b) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) to \(6 \%\) higher than the true proportion of females whose favorite color is black. (c) \(95 \%\) of random samples will produce \(95 \%\) confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance. (e) The \(95 \%\) confidence interval for \(\left(p_{\text {female }}-p_{\text {male }}\right)\) cannot be calculated with only the information given in this exercise.

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