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Chapter 10: Problem 74

The Centers for Disease Control (www.cdc.gov) periodically administers large randomized surveys to track health of Americans. In a survey of 4431 adults in \(2003 / 2004,66 \%\) were overweight (body mass index \(\mathrm{BMI} \geq 25\) ). In the most recently available survey of 5181 adults in \(2011 / 2012,69 \%\) were overweight. a. Estimate the change in the population proportion who are overweight and interpret. b. The standard error for estimating this difference equals \(0.0096 .\) What is the main factor that causes se to be so small? c. The \(95 \%\) confidence interval comparing the population proportions in \(2011 / 2012\) to the one in \(2003 / 2004\) is (0.011,0.049) . Interpret, taking into account whether 0 is in this interval.

Short Answer

Expert verified
a. The change in proportion is 3%. b. Large sample sizes make SE small. c. The interval shows a significant increase; zero is not in the interval.

Step by step solution

01

Calculate the Proportion Difference

First, calculate the change in proportion of overweight individuals from 2003/2004 to 2011/2012. The proportion from 2003/2004 is 0.66, and from 2011/2012 is 0.69. Therefore, the change in proportion is given by:\[ \Delta p = p_{2011/2012} - p_{2003/2004} = 0.69 - 0.66 = 0.03 \]This means that the proportion of overweight individuals increased by 3%.
02

Reason for Low Standard Error

The standard error (SE) for estimating the difference in proportions is small at 0.0096. One main factor contributing to this small SE is the large sample sizes ( 4431 in 2003/2004 and 5181 in 2011/2012), which reduces variability and thus decreases the SE.
03

Interpret the 95% Confidence Interval

The given 95% confidence interval for the change in population proportion is (0.011,0.049). This interval does not include 0, suggesting that we can be 95% confident there is a true increase in the proportion of overweight individuals from 2003/2004 to 2011/2012. Since 0 is not in the interval, it indicates a statistically significant increase.

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Key Concepts

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

Understanding Population Proportion
When we talk about population proportion, we refer to the fraction of the population that exhibits a particular characteristic, expressed as a percentage or a decimal. In the context of surveys like those conducted by the CDC, population proportion can indicate the percentage of adults who are overweight. For example, in 2003/2004, the population proportion of overweight individuals was 66%, or 0.66 in decimal form. Similarly, in 2011/2012, it increased to 69%, or 0.69. This change in population proportion tells us how the characteristic of interest (in this case, being overweight) has shifted over time within the population. To find the change in proportion, simply subtract the older proportion from the newer one. This gives: \[ \Delta p = p_{2011/2012} - p_{2003/2004} = 0.69 - 0.66 = 0.03 \]This 0.03, or 3%, represents the increase in the population proportion of overweight individuals over the specified period.
Explaining Standard Error
Standard error (SE) is a measure that indicates the accuracy with which a sample distribution represents a population. It is particularly useful when comparing two groups, as it quantifies the variability of the sample means. In the example with the CDC surveys, the SE for the difference in proportions is very small, at 0.0096. Such a small SE suggests that the estimate of the difference between the two proportions is very precise. A major reason for this small SE is the large sample sizes: 4431 individuals in 2003/2004 and 5181 in 2011/2012. Larger samples generally provide more accurate estimates of a population parameter because they decrease the variability of the sample means. In simple terms, the more data you collect, the more confidence you can have that your sample mean is close to the true population mean.
Importance of Statistical Significance
Statistical significance is a concept used to determine whether the results of a study are due to a real effect or are just random chance. In our example, a 95% confidence interval for the change in the proportion of overweight individuals falls between 0.011 and 0.049. The confidence interval is crucial as it helps us understand where the true population parameter is likely to lie. Since this interval does not include 0, it suggests that there is indeed a statistically significant change in the proportion of overweight individuals between 2003/2004 and 2011/2012. In practical terms, statistical significance means that we can be 95% confident that the observed increase in the proportion was not due to randomness or sampling error, but reflects a true change in the population. Thus, decisions or conclusions based on these findings have a strong backing of statistical evidence.

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Most popular questions from this chapter

Address global warming You would like to determine what students at your school would be willing to do to help address global warming and the development of alternatively fueled vehicles. To do this, you take a random sample of 100 students. One question you ask them is, "How high of a tax would you be willing to add to gasoline (per gallon) to encourage drivers to drive less or to drive more fuel-efficient cars?" You also ask, "Do you believe (yes or no) that global warming is a serious issue that requires immediate action such as the development of alternatively fueled vehicles?" In your statistical analysis, use inferential methods to compare the mean response on gasoline taxes (the first question) for those who answer yes and for those who answer no to the second question. For this analysis, a. Identify the response variable and the explanatory variable. b. Are the two groups being compared independent samples or dependent samples? Why? c. Identify a confidence interval you could form to compare the groups, specifying the parameters used in the comparison.

Vegetarians more liberal? When a sample of social science graduate students at the University of Florida gave their responses on political ideology (ranging from \(1=\) very liberal to \(7=\) very conservative \(),\) the mean was \(3.18(s=1.72)\) for the 51 nonvegetarian students and \(2.22(s=0.67)\) for the 9 vegetarian students. Software for comparing the means provides the printout, which shows results first for inferences that assume equal population standard deviations and then for inferences that allow them to be unequal. \(\begin{array}{ccclc}\text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 51 & 3.18 & 1.72 & 0.24 \\ 2 & 9 & 2.220 & 0.670 & 0.22 \\ \text { Difference } & =\mu(1)-\mu(2) & & \end{array}\) Estimate for difference: 0.960 95?a CI for difference: (-0.210,2.130) T-Test of difference \(=0\) (vs \(\neq):\) T-Value \(=1.64\) \(\mathrm{P}\) -Value \(=0.106 \mathrm{DF}=58\) Both use Pooled StDev \(=1.6162\) 95? CI for difference: (0.289,1.631) T-Test of difference \(=0(\mathrm{vs} \neq):\) T-Value \(=2.92\) P-Value \(=0.007 \quad \mathrm{DF}=30\) a. Explain why the results of the two approaches differ so much. Which do you think is more reliable? b. State your conclusion about whether the true means are plausibly equal.

A study \(^{14}\) compared personality characteristics between 49 children of alcoholics and a control group of 49 children of nonalcoholics who were matched on age and gender. On a measure of well-being, the 49 children of alcoholics had a mean of \(26.1(s=7.2)\) and the 49 subjects in the control group had a mean of \(28.8(s=6.4) .\) The difference scores between the matched subjects from the two groups had a mean of \(2.7(s=9.7)\). a. Are the groups to be compared independent samples or dependent samples? Why? b. Show all steps of a test of equality of the two population means for a two- sided alternative hypothesis. Report the P-value and interpret. c. What assumptions must you make for the inference in part b to be valid?

Risky behaviors among HIV positive female sex workers In 2014 , questionnaire surveys were administrated among 181 female sex workers in the Yunnan province of China who confirmed themselves to be HIV positive (www.ncbi.nlm. gov/pubmed/26833008). The participants were divided into two age groups -76 cases were below 35 years and 105 cases were 35 years old and above. 26 females below 35 years and 54 females of ages 35 years and above reported using drugs. Let \(p_{1}\) and \(p_{2}\) denote the population proportions of females below 35 years and of females 35 years or above who took drugs, respectively. a. Report point estimates of \(p_{1}\) and \(p_{2}\). b. Construct a \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right),\) specifying the assumptions you made to use this method. Interpret. c. Based on the interval in part b, explain why the proportion using drugs may have been quite a bit larger for females of ages 35 years or above, or it might have been only moderately larger.

From the box formula for the standard error at the end of Section 10.1 \(\frac{s e(\text { estimate } 1-\text { estimate } 2)=}{\sqrt{[s e(\text { estimate } 1)]^{2}+[s e(\text { estimate } 2)]^{2}}}\) if you know the se for each of two independent estimates, you can find the se of their difference. This is useful because often articles report an se for each sample mean or proportion, but not the se or a confidence interval for their difference. Many medical studies have used a large sample of subjects from Framingham, Massachusetts, who have been followed since 1948. A study (Annual of Internal Medicine, vol. \(138,2003,\) pp. \(24-32\) ) estimated the number of years of life lost by being obese and a smoker. For females of age 40 , adjusting for other factors, the number of years of life left were estimated to have a mean of \(46.3(s e=0.6)\) for nonsmokers of normal weight and a mean of \(33.0(s e=1.8)\) for smokers who were obese. Construct a \(95 \%\) confidence interval for the population mean number of years lost. Interpret.
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