Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Women diagnosed with breast cancer whose tumors have not spread may be faced with a decision between two surgical treatments - mastectomy (removal of the breast) or lumpectomy (only the tumor is removed). In a long-term study of the effectiveness of these two treatments, 701 women with breast cancer were randomly assigned to one of two treatment groups. One group received mastectomies and the other group received lumpectomies and radiation. Both groups were followed for 20 years after surgery. It was reported that there was no statistically significant difference in the proportion surviving for 20 years for the two treatments (Associated Press, October 17,2002 ). What hypotheses do you think the researchers tested in order to reach the given conclusion? Did the researchers reject or fail to reject the null hypothesis?

Short Answer

Expert verified
The researchers tested the Null Hypothesis that states no significant difference in the 20-year survival rate for patients who received mastectomies or lumpectomies and radiation. Since it was reported that there was no statistically significant difference in the proportions surviving for 20 years for the two treatments, this suggests that the researchers failed to reject the Null Hypothesis.

Step by step solution

01

Understanding the Null Hypothesis

The Null Hypothesis \(H_0\) for this study states that there is no significant difference in the survival proportion between the two groups. This means the proportion of women who survive for 20 years after being treated with mastectomy is the same as the proportion of women who survive for 20 years after being treated with lumpectomy and radiation.
02

Understanding the Alternative Hypothesis

The Alternative Hypothesis \(H_1\) for this study states that there is a significant difference in the survival proportion between the two groups. This means the proportion of women who survive for 20 years after being treated with mastectomy is not the same as the proportion of women who survive for 20 years after being treated with lumpectomy and radiation.
03

Interpreting the Conclusion of the researchers

The researchers did not find a statistically significant difference hence, 'Failed to reject the Null Hypothesis'. This means according to the data collected and analyzed, there isn't enough evidence to suggest that one treatment is significantly better than the other in terms of the 20-year survival rate.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), forms the cornerstone of scientific inquiry. It suggests that there is no effect or no difference in the context of the variables being tested. In the case of the study on breast cancer treatments, the null hypothesis proposes that there is no statistically significant difference in the 20-year survival rates between women who underwent mastectomies and those who received lumpectomies with radiation.

For a study like this, the null hypothesis acts as a default assumption that researchers aim to test. It's comparable to a presumption of innocence in a court case; the null hypothesis is innocent until proven guilty by the data. If the evidence (data) does not strongly contradict this hypothesis, we continue to assume it's true.
Alternative Hypothesis
The alternative hypothesis, represented as \( H_1 \), is the statement that contradicts the null hypothesis. It is what researchers hope to find evidence for, essentially positing that a genuine effect or difference exists. In this breast cancer treatment study, the alternative hypothesis would assert that there is a significant difference in the 20-year survival rates of breast cancer patients between the two treatments: mastectomy and lumpectomy combined with radiation.

If the null hypothesis is like a status quo, the alternative hypothesis is the challenger. Proving this hypothesis typically means that there is sufficient evidence to suggest a variable change or a noticeable effect. In our example, a significant finding would involve determining that one surgical treatment leads to a better long-term survival outcome than the other.
Statistical Significance
Statistical significance is a term used to indicate whether the observed effect in a study justifies rejecting the null hypothesis. When researchers find a result statistically significant, it means the data suggests there is a low probability the observed effect is due to chance alone. Often, a significance level (denoted by \( \alpha\)) of 0.05 is used as a standard threshold. This signifies that there is a 5% probability that the results occurred by random chance rather than a true effect.

In the breast cancer study, the researchers concluded that there was "no statistically significant difference" in the survival rates between the two treatment groups. This indicates that the data did not provide strong enough evidence against the null hypothesis, hence the researchers failed to reject it. It doesn't prove that the null hypothesis is true, but rather that there isn't enough evidence to declare that it's false.
Survival Analysis
Survival analysis is a statistical method used to analyze data in which the outcome variable is the time until an event occurs. It's particularly useful when studying the effectiveness of treatments or interventions over time. This method accounts not only for whether a patient survived a treatment period, but also how long they survived after the treatment.

The breast cancer treatment study is a classic example of survival analysis, focusing on the long-term survival of patients after undergoing different surgical procedures. The key statistic of interest is the survival probability at a certain time point, in this case, 20 years. By using survival analysis, researchers could comprehensively compare the efficacy of the two treatment options over time, providing a more nuanced understanding of their impact on patient longevity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Acrylic bone cement is commonly used in total joint replacement to secure the artificial joint. Data on the force (measured in Newtons, \(N\) ) required to break a cement bond under two different temperature conditions and in two different mediums appear in the accompanying table. (These data are consistent with summary quantities appearing in the paper "Validation of the Small-Punch Test as a Technique for Characterizing the Mechanical Properties of Acrylic Bone Cement" (Journal of Engineering in Medicine [2006]: 11-21).) $$ \begin{array}{lcl} \text { Temperature } & \text { Medium } & \text { Data on Breaking Force } \\\ \hline \text { 22 degrees } & \text { Dry } & 100.8,141.9,194.8,118.4, \\ & & 176.1,213.1 \\ \text { 37 degrees } & \text { Dry } & 302.1,339.2,288.8,306.8, \\ & & 305.2,327.5 \\ \text { 22 degrees } & \text { Wet } & 385.3,368.3,322.6,307.4, \\ & & 357.9,321.4 \\ \text { 37 degrees } & \text { Wet } & 363.5,377.7,327.7,331.9, \\ & & 338.1,394.6 \\ \hline \end{array} $$ a. Estimate the difference between the mean breaking force in a dry medium at 37 degrees and the mean breaking force at the same temperature in a wet medium using a \(90 \%\) confidence interval. b. Is there sufficient evidence to conclude that the mean breaking force in a dry medium at the higher temperature is greater than the mean breaking force at the lower temperature by more than \(100 N ?\) Test the relevant hypotheses using a significance level of .10 .

After the 2010 earthquake in Haiti, many charitable organizations conducted fundraising campaigns to raise money for emergency relief. Some of these campaigns allowed people to donate by sending a text message using a cell phone to have the donated amount added to their cell-phone bill. The report "Early Signals on Mobile Philanthropy: Is Haiti the Tipping Point?" (Edge Research, 2010 ) describes the results of a national survey of 1526 people that investigated the ways in which people made donations to the Haiti relief effort. The report states that \(17 \%\) of Gen \(Y\) respondents (those born between 1980 and 1988 ) and \(14 \%\) of Gen \(X\) respondents (those born between 1968 and 1979 ) said that they had made a donation to the Haiti relief effort via text message. The percentage making a donation via text message was much lower for older respondents. The report did not say how many respondents were in the Gen \(\mathrm{Y}\) and Gen \(\mathrm{X}\) samples, but for purposes of this exercise, suppose that both sample sizes were 400 and that it is reasonable to regard the samples as representative of the Gen \(\mathrm{Y}\) and Gen \(\mathrm{X}\) populations. a. Is there convincing evidence that the proportion of those in Gen Y who donated to Haiti relief via text message is greater than the proportion for Gen X? Use \(\alpha=.01\). b. Estimate the difference between the proportion of Gen \(\mathrm{Y}\) and the proportion of Gen \(\mathrm{X}\) that made a donation via text message using a \(99 \%\) confidence interval. Provide an interpretation of both the interval and the associated confidence level.

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.

Do certain behaviors result in a severe drain on energy resources because a great deal of energy is expended in comparison to energy intake? The article "The Energetic Cost of Courtship and Aggression in a Plethodontid Salamander" (Ecology [1983]: 979-983) reported on one of the few studies concerned with behavior and energy expenditure. The accompanying table gives oxygen consumption \((\mathrm{mL} / \mathrm{g} / \mathrm{hr})\) for male-female salamander pairs. (The determination of consumption values is rather complicated. It is partly for this reason that so few studies of this type have been carried out.) $$ \begin{array}{lccc} \text { Behavior } & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Sample } \\ \text { Mean } \end{array} & \begin{array}{c} \text { Sample } \\ \text { sd } \end{array} \\ \hline \text { Noncourting } & 11 & .072 & .0066 \\ \text { Courting } & 15 & .099 & .0071 \\ \hline \end{array} $$ a. The pooled \(t\) test is a test procedure for testing \(H_{0}: \mu_{1}-\mu_{2}=\) hypothesized value when it is reasonable to assume that the two population distributions are normal with equal standard deviations \(\left(\sigma_{1}=\right.\) \(\sigma_{2}\) ). The test statistic for the pooled \(t\) test is obtained by replacing both \(s_{1}\) and \(s_{2}\) in the two-sample \(t\) test statistic with \(s_{p}\) where \(s_{p}=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}\) When the population distributions are normal with equal standard deviations and \(H_{0}\) is true, the resulting pooled \(t\) statistic has a \(t\) distribution with \(\mathrm{df}=n_{1}+\) \(n_{2}-2 .\) For the reported data, the two sample standard deviations are similar. Use the pooled \(t\) test with \(\alpha=.05\) to determine whether the mean oxygen consumption for courting pairs is higher than the mean oxygen consumption for noncourting pairs. b. Would the conclusion in Part (a) have been different if the two-sample \(t\) test had been used rather than the pooled \(t\) test?

The article "Fish Oil Staves Off Schizophrenia" (USA Today, February 2, \(2 \mathrm{O} 1 \mathrm{O}\) ) describes a study in which 81 patients age 13 to 25 who were considered atrisk for mental illness were randomly assigned to one of two groups. Those in one group took four fish oil capsules daily. The other group took a placebo. After 1 year, \(5 \%\) of those in the fish oil group and \(28 \%\) of those in the placebo group had become psychotic. Is it appropriate to use the two-sample \(z\) test of this section to test hypotheses about the difference in the proportions of patients receiving the fish oil and the placebo treatments who became psychotic? Explain why or why not.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free