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Women's satisfaction with appearance A special issue of Newsweek in March 1999 on women and their health reported results of a poll of 757 American women aged 18 or older. When asked, "How satisfied are you with your overall physical appearance?" \(30 \%\) said very satisfied, \(54 \%\) said somewhat satisfied, \(13 \%\) said not too satisfied, and \(3 \%\) said not at all satisfied. True or false: Since all these percentages are based on the same sample size, they all have the same margin of error.

Short Answer

Expert verified
False: Different satisfaction categories have different margins of error.

Step by step solution

01

Understanding Margin of Error

The margin of error depends on both the sample size and the proportion of the population that chooses a particular answer. Although our sample size remains constant at 757 women, the proportion for each satisfaction level differs, which affects the margin of error calculations.
02

Proportion Calculation

For each satisfaction category, let's set up the proportion: - Very satisfied: Proportion = 0.30 - Somewhat satisfied: Proportion = 0.54 - Not too satisfied: Proportion = 0.13 - Not at all satisfied: Proportion = 0.03 These proportions represent the percentage of women who chose each category.
03

Margin of Error Formula

The margin of error for a proportion can be calculated using the formula: \[E = z \times \sqrt{\frac{p(1-p)}{n}}\]where: - \( z \) is the z-score for the desired confidence level, typically 1.96 for a 95% confidence interval,- \( p \) represents the proportion,- \( n \) is the sample size, here 757.
04

Calculation of Margin of Error

Substitute each category’s proportion into the margin of error formula:- Very satisfied: \( E = 1.96 \times \sqrt{\frac{0.30 \times 0.70}{757}} \)- Somewhat satisfied: \( E = 1.96 \times \sqrt{\frac{0.54 \times 0.46}{757}} \)- Not too satisfied: \( E = 1.96 \times \sqrt{\frac{0.13 \times 0.87}{757}} \)- Not at all satisfied: \( E = 1.96 \times \sqrt{\frac{0.03 \times 0.97}{757}} \) Each calculation will yield a different margin of error.
05

Conclusion

Since the margin of error is dependent on the specific proportion \( p \), each satisfaction category will have a different margin of error owing to the varying proportions, despite the sample size being the same.

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Key Concepts

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

Sampling Error
Sampling error occurs when we use a sample of the population to draw conclusions about the entire population. This discrepancy arises because a sample, no matter how large, may not perfectly represent the larger group.
In the context of the Newsweek poll of 757 American women, sampling error refers to the potential difference between the poll results and the actual opinions of all American women of that demographic.
The sample consisted of women aged 18 or older, and we assume that it was randomly selected to ensure equal chances of representation. However, random chance could still result in a sample that is not a perfect mirror of the population.
Good sampling practices minimize sampling error, though it can never be fully eliminated.
Confidence Interval
A confidence interval provides a range within which we expect the true population parameter to lie. When we report a poll result like "30% very satisfied," we often also report a confidence interval, which might look something like "30% ± 4%".
This interval indicates that we are fairly certain (often 95% confident) that the true proportion of all American women who are very satisfied lies between 26% and 34%.
The width of a confidence interval is affected by the sample size and the variation within the sample. A larger sample size generally results in a narrower confidence interval, signaling more precise estimates.
Confidence intervals use statistical tools like the z-score to account for the desired level of confidence (e.g., 95% confidence uses a z-score of 1.96). By being transparent about the range, confidence intervals provide insight into the reliability of the survey results.
Proportion Calculation
Proportion calculation is crucial when analyzing survey data to understand how many respondents fall into specific categories. In the Newsweek poll, proportions translate to the percentage of respondents in each satisfaction level. For instance, 30% very satisfied means 0.30 in proportion terms.
Calculating proportions is a straightforward but key process: take the number of respondents in a category and divide by the total sample size. Proportions help convey the share of the sample that supports a particular view.
Using proportions allows for calculations of other metrics, such as the margin of error. Without accurate proportion calculations, other statistical computations could be skewed or inaccurate.
Survey Analysis
Survey analysis involves examining survey results to glean insights about the surveyed population's preferences and tendencies. It often includes categorizing data, computing relevant statistics, and interpreting these in context.
The Newsweek poll provides raw numbers of women occupying each satisfaction category along with the calculated percentages. Understanding these numbers helps ascertain general trends, such as a majority of women feeling somewhat satisfied with their appearance.
Proper survey analysis also involves scrutinizing the methodology used, ensuring the sample was well-chosen, and interpreting results in light of the margin of error and confidence intervals.
Through careful analysis, surveys like these can effectively inform public opinion, strategic marketing, or policy-making.

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Most popular questions from this chapter

Multiple choice: CI property Increasing the confidence level causes the margin of error of a confidence interval to \((\) a) increase, \((b)\) decrease, \((c)\) stay the same.

When the GSS recently asked subjects whether it should or should not be the government's responsibility to impose strict laws to make industry do less damage to the environment (variable GRNLAWS), 1403 of 1497 subjects said yes. a. What assumptions are made to construct a \(95 \%\) confidence interval for the population proportion who would say yes? Do they seem satisfied here? b. Construct the \(95 \%\) confidence interval. Interpret in context. Can you conclude whether or not a majority or minority of the population would answer yes?

Nutrient effect on growth rate Researchers are interested in the effect of a certain nutrient on the growth rate of plant seedlings. Using a hydroponics grow procedure that utilized water containing the nutrient, they planted six tomato plants and recorded the heights of each plant 14 days after germination. Those heights, measured in millimeters, were as follows: \(55.5,60.3,60.6,62.1,65.5,69.2 .\) a. Find a point estimate of the population mean height of this variety of seedling 14 days after germination. b. A method that we'll study in Section 8.3 provides a margin of error of \(4.9 \mathrm{~mm}\) for a \(95 \%\) confidence interval for the population mean height. Construct that interval. c. Use this example to explain why a point estimate alone is usually insufficient for statistical inference.

South Africa study The researcher planning the study in South Africa also will estimate the population proportion having at least a high school education. No information is available about its value. How large a sample size is needed to estimate it to within 0.07 with \(95 \%\) confidence?

Anorexia in teenage girls A study \(^{6}\) compared various therapies for teenage girls suffering from anorexia, an eating disorder. For each girl, weight was measured before and after a fixed period of treatment. The variable measured was the change in weight, \(X=\) weight at the end of the study minus weight at the beginning of the study. The therapies were designed to aid weight gain, corresponding to positive values of \(X .\) For the sample of 17 girls receiving the family therapy, the changes in weight during the study were 11,11,6,9,14,-3,0,7,22,-5,-4,13,13,9,4,6,11 a. Plot these with a dot plot or box plot, and summarize. b. Using a calculator or software, show that the weight changes have \(\bar{x}=7.29\) and \(s=7.18\) pounds. c. Using a calculator or software, show that the standard error of the sample mean was se \(=1.74\). d. To use the \(t\) distribution, explain why the \(95 \%\) confidence interval uses the \(t\) -score equal to 2.120 . e. Let \(\mu\) denote the population mean change in weight for this therapy. Using results from parts \(\mathrm{b}, \mathrm{c},\) and \(\mathrm{d}\), show that the \(95 \%\) confidence interval for \(\mu\) is \((3.6,11.0) .\) Explain why this suggests that the true mean change in weight is positive, but possibly quite small.

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