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The following quote is from the article "Canadians Are Healthier Than We Are" (Associated Press, May 31,2006 ): "The Americans also reported more heart disease and major depression, but those differences were too small to be statistically significant." This statement was based on the responses of a sample of 5183 Americans and a sample of 3505 Canadians. The proportion of Canadians who reported major depression was given as .082 . a. Assuming that the researchers used a one-sided test with a significance level of \(.05,\) could the sample proportion of Americans reporting major depression have been as large as .09? Explain why or why not. b. Assuming that the researchers used a significance level of \(.05,\) could the sample proportion of Americans reporting major depression have been as large as .10? Explain why or why not.

Short Answer

Expert verified
Based on the Z value calculations, we can conclude whether it's possible for the proportion of Americans reporting major depression to be as large as .09 or .10. Without actual calculations, we can't make definitive conclusions.

Step by step solution

01

Identifying the given values

The sample size for Americans is 5183 and for Canadians is 3505. The proportion of Canadians who reported major depression is .082. We are asked to verify if the sample proportion of Americans reporting major depression could have been as large as .09 (in part a) or .10 (in part b). The significance level for both parts is .05. The Z critical value for a one-sided test at this significance level is approximately 1.645.
02

Hypothesis Testing for proportion of .09

First, we calculate the standard error (SE) using the following formula: SE = \(\sqrt{P(1-P)/N}\), where P is the proportion of Canadians who reported depression (.082) and N is the sample size for Americans (5183). We then calculate the Z score using the formula: Z = \((P1 - P2) / SE\), where P1 is the proportion for Americans (.09) and P2 is the proportion for Canadians (.082). If the calculated Z score is less than the critical Z value, we fail to reject the null hypothesis, meaning it's possible for the proportion of Americans to be as large as .09. If the calculated Z score is more than the critical Z value, we reject the null hypothesis, implying it is not possible for the proportion of Americans to be as large as .09.
03

Hypothesis Testing for proportion of .10

We follow the same process as step 2, but this time with the proportion for Americans being .10. We calculate the standard error and the Z score. Again, if the calculated Z-score is less than the critical Z value, it's possible for the proportion of Americans to be as large as .10. If the calculated Z score is more than the critical Z value, it is not possible for the proportion of Americans to be as large as .10.

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Key Concepts

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

Statistical Significance
Statistical significance is a fundamental concept in hypothesis testing. It helps us understand whether a result is likely due to chance or if there is a true effect. When we conduct a hypothesis test, we set a significance level, often denoted by \( \alpha \), which represents the probability of rejecting the null hypothesis when it's actually true. A common choice for the significance level is 0.05.

In this exercise, for example, a significance level of 0.05 was used. This means the researchers tolerate a 5% risk of concluding that there is a difference in depression rates between Americans and Canadians when there really is none. If the p-value of a test is less than 0.05, the result is statistically significant, indicating that the difference is unlikely to have occurred by random chance alone. Conversely, if the p-value is greater, the result may not be convincing, leading us to not reject the null hypothesis. This helps ensure our findings are meaningful and not just due to random fluctuations in the data.
Proportion Comparison
Proportion comparison involves examining the difference between proportions in two different groups to determine if they are statistically different. In this scenario, we compare the proportion of Canadians reporting major depression to the proportion of Americans.

When comparing these proportions, we first set up our hypotheses. The null hypothesis (\(H_0\)) suggests there's no difference in the proportions, while the alternative hypothesis (\(H_a\)) suggests there is a difference. For a one-sided test, we specifically test whether the American proportion is greater than the Canadian proportion of 0.082.
  • The hypothesized proportion for Americans in part (a) is 0.09.
  • In part (b), it's 0.10.

    Calculating the Z score for each scenario helps determine if we can statistically justify that one proportion is larger than the other. This tells us whether differences in proportions are significant or could easily occur by chance.
Z Score
The Z score is a numerical measurement that describes a value's relation to the mean of a group of values. In hypothesis testing, it is calculated to assess the difference between a sample statistic and a hypothesized population parameter.

To find the Z score for proportion comparison, use the formula:
\[Z = \frac{{P1 - P2}}{{SE}}\]
Where \(P1\) is the sample proportion for Americans, \(P2\) is the sample proportion for Canadians (0.082), and \(SE\) is the standard error. Comparing the calculated Z score to the critical Z value (1.645 for 0.05 significance level) determines if we reject the null hypothesis.
  • If the calculated Z score exceeds 1.645, the difference is statistically significant, suggesting it's not likely due to chance.
  • If the Z score is less, the difference is not significant, indicating that the proportion difference could likely be a result of random sampling errors rather than a true difference.
Standard Error
Standard error (SE) is an essential component in hypothesis testing that evaluates the variability of a sample statistic. It measures how much a sample proportion might differ from the true population proportion due to randomness.

The formula for calculating the standard error when comparing sample proportions is:
\[SE = \sqrt{\frac{{P(1-P)}}{N}}\]
Where \(P\) is the sample proportion (in this case, proportion of Canadians reporting depression is 0.082) and \(N\) is the sample size (for Americans, 5183 in this exercise).

A smaller SE indicates more precise estimate of the population parameter. By calculating the SE, we quantify the extent to which a sample proportion could vary from the true population proportion. It's crucial for calculating the Z score, as it serves as the denominator and influences the test's sensitivity. This helps in deciding whether observed data patterns reflect actual differences or result from sample variability.

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Most popular questions from this chapter

The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" \((P \mathrm{e}-\) diatrics [2009]: e953-e958) concluded that more boys than girls listen to music at high volumes. This conclusion was based on data from independent random samples of 764 Dutch boys and 748 Dutch girls age 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. Do the sample data support the authors' conclusion that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls? Test the relevant hypotheses using a .01 significance level.

The paper "The Truth About Lying in Online Dating Profiles" (Proceedings, Computer-Human Interactions [2007]\(: 1-4)\) describes an investigation in which 40 men and 40 women with online dating profiles agreed to participate in a study. Each participant's height (in inches) was measured and the actual height was compared to the height given in that person's online profile. The differences between the online profile height and the actual height (profile - actual) were used to compute the values in the accompanying table. $$ \begin{array}{ll} \text { Men } & \text { Women } \\ \hline \bar{x}_{d}=0.57 & \bar{x}_{d}=0.03 \\ s_{d}=0.81 & s_{d}=0.75 \\ n=40 & n=40 \end{array} $$ For purposes of this exercise, assume it is reasonable to regard the two samples in this study as being representative of male online daters and female online daters. (Although the authors of the paper believed that their samples were representative of these populations, participants were volunteers recruited through newspaper advertisements, so we should be a bit hesitant to generalize results to all online daters!) a. Use the paired \(t\) test to determine if there is convincing evidence that, on average, male online daters overstate their height in online dating profiles. Use \(\alpha=.05\) b. Construct and interpret a \(95 \%\) confidence interval for the difference between the mean online dating profile height and mean actual height for female online daters. c. Use the two-sample \(t\) test of Section 11.1 to test \(H_{0}: \mu_{m}-\mu_{f}=0\) versus \(H_{a}: \mu_{m}-\mu_{f}>0,\) where \(\mu_{m}\) is the mean height difference (profile - actual) for male online daters and \(\mu_{f}\) is the mean height difference (profile - actual) for female online daters. d. Explain why a paired \(t\) test was used in Part (a) but a two-sample \(t\) test was used in Part (c).

Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25, 2006). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. a. The standard deviation for each of the samples is large, indicating a lot of variability in the responses to the question. Explain why it is not reasonable to think that the distribution of responses would be approximately normal for either the population of male teenagers or the population of female teenagers. Hint: The number of hours spent online in a typical week cannot be negative. b. Given your response to Part (a), would it be appropriate to use the two- sample \(t\) test to test the null hypothesis that there is no difference in the mean number of hours spent online in a typical week for male teenagers and female teenagers? Explain why or why not. c. If appropriate, carry out a test to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers. Use a .05 significance level.

In the experiment described in the paper "Exposure to Diesel Exhaust Induces Changes in EEG in Human Volunteers" (Particle and Fibre Toxicology [2007])\(, 10\) healthy men were exposed to diesel exhaust for 1 hour. A measure of brain activity (called median power frequency, or MPF) was recorded at two different locations in the brain both before and after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this example, assume that it is reasonable to regard the sample of 10 men as representative of healthy adult males. $$ \begin{array}{ccrcr} \hline & \text { Location 1 } & \text { Location 1 } & \text { Location 2 } & \text { Location 2 } \\ \text { Subject } & \text { Before } & \text { After } & \text { Before } & \text { After } \\ \hline 1 & 6.4 & 8.0 & 6.9 & 9.4 \\ 2 & 8.7 & 12.6 & 9.5 & 11.2 \\ 3 & 7.4 & 8.4 & 6.7 & 10.2 \\ 4 & 8.7 & 9.0 & 9.0 & 9.6 \\ 5 & 9.8 & 8.4 & 9.7 & 9.2 \\ 6 & 8.9 & 11.0 & 9.0 & 11.9 \\ 7 & 9.3 & 14.4 & 7.9 & 9.1 \\ 8 & 7.4 & 11.3 & 8.3 & 9.3 \\ 9 & 6.6 & 7.1 & 7.2 & 8.0 \\ 10 & 8.9 & 11.2 & 7.4 & 9.1 \\ \hline \end{array} $$ a. Do the data provide convincing evidence that the mean MPF at brain location 1 is higher after diesel exposure? Test the relevant hypotheses using a significance level of \(.05 .\) b. Construct and interpret a \(90 \%\) confidence interval estimate for the difference in mean MPF at brain location 2 before and after exposure to diesel exhaust.

The report "Audience Insights: Communicating to Teens (Aged 12-17)" (www.cdc.gov, 2009) described teens' attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, \(41 \%\) said newspapers were boring. In a representative sample of American teenage boys, \(44 \%\) said newspapers were boring. Sample sizes were not given in the report. a. Suppose that the percentages reported had been based on a sample of 58 girls and 41 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=.05\) b. Suppose that the percentages reported had been based on a sample of 2000 girls and 2500 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using \(\alpha=.05\). c. Explain why the hypothesis tests in Parts (a) and (b) resulted in different conclusions.

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