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Chapter 10: Problem 123

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.

Short Answer

Expert verified
The 95% confidence interval for the difference in mean years is (9.58, 17.02).

Step by step solution

01

Identify Given Values

We are given two estimates with their standard errors: the mean number of years for nonsmokers of normal weight is 46.3 with a standard error of 0.6, and for smokers who are obese, it is 33.0 with a standard error of 1.8.
02

Calculate the Standard Error of the Difference

To find the standard error of the difference between these means, use the formula: \[se(\text{difference}) = \sqrt{[se(\text{estimate 1})]^2 + [se(\text{estimate 2})]^2} = \sqrt{0.6^2 + 1.8^2}\]Calculating this gives:\[se(\text{difference}) = \sqrt{0.36 + 3.24} = \sqrt{3.6} \approx 1.897\]
03

Determine Mean Difference

Subtract the mean of smokers who are obese from the mean of nonsmokers of normal weight:\[46.3 - 33.0 = 13.3\]
04

Calculate the 95% Confidence Interval

Use the formula: \[\text{Mean Difference} \pm Z \times se(\text{difference})\]where \(Z\) is the Z-score for a 95% confidence level, typically 1.96.\[13.3 \pm 1.96 \times 1.897 \]Calculate the margin of error:\[\text{Margin of Error} = 1.96 \times 1.897 \approx 3.720\]Therefore, the confidence interval is:\[13.3 \pm 3.720 = (9.58, 17.02)\]
05

Interpretation of the Confidence Interval

The 95% confidence interval for the difference in mean years of life left is (9.58, 17.02). This means we are 95% confident that the true mean difference in the number of years of life left between nonsmokers of normal weight and smokers who are obese falls within this interval.

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Key Concepts

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

Standard Error
Understanding the concept of 'Standard Error' is crucial when analyzing data in various fields. The standard error (SE) measures how much the estimated value from a sample is expected to deviate from the true population parameter. It essentially reflects the degree of uncertainty or error of the sample's estimate. The smaller the standard error, the more precise the estimate is.

In the context of the exercise, the SE is used to assess the variation in the mean number of years of life left. For nonsmokers of normal weight, the SE is 0.6. For obese smokers, it is 1.8. These individual SE values help us gauge how close our sample estimates might be to the true population values.

It's important to note that the standard error is influenced by the sample size. Larger sample sizes typically result in a smaller SE, indicating more reliable estimates. To find the SE of the difference between two sample estimates, as shown in the exercise solution, we use the formula:
  • \[ se(\text{difference}) = \sqrt{[se(\text{estimate 1})]^2 + [se(\text{estimate 2})]^2} \]
This formula helps us determine the SE of the mean difference, leveraging the SE values from each independent estimate within the samples.
Mean Difference
The 'Mean Difference' is a central concept when comparing two groups in a study. It represents the average difference between the two means of interest.

In our exercise, we are looking at the mean number of years of life left for two groups: nonsmokers of normal weight and smokers who are obese. By subtracting the mean life expectancy of smokers who are obese (33.0 years) from that of nonsmokers of normal weight (46.3 years), we calculate the mean difference:

  • \[ 46.3 - 33.0 = 13.3 \]
This mean difference of 13.3 years gives us a point estimate of the difference in life expectancy between the two groups.

Calculating the mean difference alone, however, does not convey the full picture. It's crucial to also measure the precision of this estimate by looking at the standard error of the difference and constructing a confidence interval. This helps us understand how much the mean difference might vary from the true difference in the population.
Z-Score
The term 'Z-Score' refers to a statistical measure that describes a value's position relative to the mean of a group of values, expressed in terms of standard deviations. This concept is widely used in hypothesis testing and the construction of confidence intervals.

In our exercise, the Z-score plays a crucial role when establishing a confidence interval. For example, when we want to determine a 95% confidence interval for the mean difference, we select a Z-score of 1.96. This Z-score corresponds to a standard normal distribution, covering approximately 95% of the data.

  • The confidence interval is calculated as: \[ \text{Mean Difference} \pm Z \times se(\text{difference}) \]
In our case, it resulted in:
  • \[ 13.3 \pm 1.96 \times 1.897 = (9.58, 17.02) \]
This confidence interval provides a range in which we are 95% certain that the true mean difference in life expectancy for the populations lies. By understanding the role of Z-scores, we can better assess statistical inferences and make stronger data-driven conclusions.

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Most popular questions from this chapter

Suppose that a \(99 \%\) confidence interval for the difference \(p_{1}-p_{2}\) between the proportions of men and women in California who are alcoholics equals \((0.02,0.09) .\) Choose the best correct choice. a. We are \(99 \%\) confident that the proportion of alcoholics is between 0.02 and 0.09 . b. We are \(99 \%\) confident that the proportion of men in California who are alcoholics is between 0.02 and 0.09 larger than the proportion of women in California who are. c. We can conclude that the population proportions may be equal. d. We are \(99 \%\) confident that a minority of California residents are alcoholics. e. Since the confidence interval does not contain \(0,\) it is impossible that \(p_{1}=p_{2}\)

A Danish study of individuals born at a Copenhagen hospital between 1959 and 1961 reported higher mean IQ scores for adults who were breast-fed for longer lengths of time as babies (E. Mortensen et al., \(J A M A,\) vol. \(287,2002,\) pp. \(2365-2371\) ). The mean IQ score was \(98.1(s=15.9)\) for the \(272 \mathrm{sub}\) jects who had been breast-fed for no longer than a month and \(108.2(s=13.1)\) for the 104 subjects who had been breast-fed for seven to nine months. a. With software that can analyze summarized data, use an inferential method to analyze whether the corresponding population means differ, assuming the population standard deviations are equal. Interpret. b. Was this an experimental or an observational study? Can you think of a potential lurking variable?

Kidnapping in southern and eastern European countries The following data on kidnapping offences in countries of east and south Europe in 2014 were obtained from https://data.unodc.org. (Crime and Criminal Justice \(\rightarrow>\) Crime \(\rightarrow>\) Kidnapping \(\rightarrow\) Filter by Region and Sub Region as appropriate) \(\begin{array}{ll}\text { Eastern Europe: } & 31,95,12,3,292,88,369,10\end{array}\) Southern Europe: \(\quad 2,1,3,1,58,297,22,376,11,5,99,8\) Using statistical software, a. Construct and interpret a plot comparing responses by region. Kidnapping in southern and eastern European countries The following data on kidnapping offences in countries of east and south Europe in 2014 were obtained from https://data.unodc.org. (Crime and Criminal Justice \(->\) Crime \(\rightarrow>\) Kidnapping \(->\) Filter by Region and Sub Region as appropriate) Eastern Europe: 31,95,12,3,292,88,369,10 Southern Europe: 2,1,3,1,58,297,22,376,11,5,99,8 Using statistical software, a. Construct and interpret a plot comparing responses by region.

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.

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