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In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: \(2024-2033\) ). The following statement appeared in the paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004)\) a. What hypotheses did the researchers test to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(\quad t=5.00\) iii. \(t=2.33\) ii. \(t=2.65\) iv. \(t=1.47\)

Short Answer

Expert verified
a. The hypotheses the researchers were testing are: \(H0: \mu1 = \mu2\) and \(H1: \mu1 \neq \mu2\).\n b. Considering the provided options, the most probable test statistic for the hypothesis test could be \(t=5.00\) due to sample size and P-value.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (H0) is that the means of the two samples (claims without medical errors and claims with errors) are equal. The alternative hypothesis (H1) is that the means of the two samples are not equal. In mathematical form: \(H0: \mu1 = \mu2\) \(H1: \mu1 \neq \mu2\) where \(\mu1\) is the mean settlement of claims without errors and \(\mu2\) is the mean settlement for claims with errors.
02

Evaluate the Possible t-values

Given the alternative hypothesis is two-tailed (means not equal), the t-value could be either positive or negative, depending on whether \(\mu1\) is larger or smaller than \(\mu2\). Moreover, since the P-value given is 0.004, this means that the absolute t-value must be quite large to be in the extreme 0.4% of the t-distribution. Looking at the options provided, \(t=5.00\) and \(t=2.65\) are the largest values and hence the most probable t-statistics.
03

Provide Reasoning for Selections

While both \(t=5.00\) and \(t=2.65\) are plausible, typically a t-value that results in a P-value of 0.004, with large sample sizes (over 400), would likely be somewhat larger, thus \(t=5.00\) would be the most likely candidate. This is not a certainty, however, without the degrees of freedom and the exact t-distribution that the researchers used.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about population parameters based on sample data. In the given malpractice claims study, the researchers used hypothesis testing to determine if there was a significant difference between the average payments of claims with medical errors and those without. This process begins by formulating two contrasting statements: the null hypothesis and the alternative hypothesis. The null hypothesis posits that there is no effect or no difference, whereas the alternative hypothesis suggests that there is an effect or a difference. In our example, the null hypothesis is that the average payments for both groups of claims are equal, while the alternative hypothesis contends that they are not.

The validity of the null hypothesis is tested using a chosen significance level and a corresponding test statistic. If the test statistic falls into a critical region, which is determined by the significance level, the null hypothesis is rejected; otherwise, it is not rejected. This process allows researchers to draw conclusions with a known level of uncertainty, which is controlled by the probability of making a Type I error (incorrectly rejecting a true null hypothesis).

Hypothesis testing is fundamental in research as it provides a structured way to infer about populations based on random samples, giving it a broad range of applications across various fields.
T-Value
The t-value, often called the t-statistic, is an essential concept in hypothesis testing, specifically when the standard deviation of a population is unknown and the sample size is small (typically less than 30). It is calculated as the difference between the sample mean and the hypothesized population mean divided by the standard error of the sample mean.

The formula for the t-statistic is: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, 's' is the sample standard deviation, and 'n' is the sample size. In the context of the malpractice claims study, a t-value is used to determine whether the difference between the mean settlements of the two groups is statistically significant. The farther the t-value is from zero, the less likely it is that the observed sample mean difference is due to random chance, thereby providing evidence against the null hypothesis.
Null Hypothesis
The null hypothesis (\( H_0 \) is a statement of no effect, no difference, or no change. It is the starting point for hypothesis testing and serves as the default assumption that there is no significant relationship between the variables being tested. The null hypothesis is what the researcher aims to refute or disprove.

In our exercise, the null hypothesis claims that the average payouts for malpractice claims not involving medical errors (\( \mu_1 \) and those involving errors (\( \mu_2 \) are the same (\( H_0: \mu_1 = \mu_2 \). This hypothesis is formulated with the notion that any observed difference in the sample averages could be due to random variation or sampling error. To reject the null hypothesis, we require sufficient evidence from our sample data that suggests a significant difference exists between the two populations in question.
Alternative Hypothesis
The alternative hypothesis (\( H_A \) or \( H_1 \) is the hypothesis that the researcher wants to support. It directly contradicts the null hypothesis and is considered to be true if the null hypothesis is rejected. It represents a new theory or belief based on the evidence available from the sample data.

In a two-tailed test, such as the one in our malpractice claims example, the alternative hypothesis is that there is a difference in the mean settlements between the two groups: \( H_A: \mu_1 eq \mu_2 \). The alternative hypothesis can also be one-tailed if the research is only interested in whether one mean is greater or less than the other. Since the researchers in the study indicated a significant difference in settlements between claims, they were specifically looking to find evidence to support this alternative hypothesis. Successful support for the alternative hypothesis leads to new insights and developments in the field of study.
P-Value
The P-value is a critical component in hypothesis testing used to measure the strength of evidence against the null hypothesis. It represents the probability of obtaining a test statistic at least as extreme as the one observed, given that the null hypothesis is true. In simple terms, it quantifies how surprising the observed data is under the assumption that the null hypothesis holds.

A smaller P-value indicates stronger evidence against the null hypothesis. In the case of the malpractice claims study, the P-value of 0.004 suggests a very low probability that the observed difference in settlements happened by chance, hence providing strong evidence against the null hypothesis. Typically, researchers decide upon a significance level (commonly 0.05) before the test, and if the P-value is below this threshold, they reject the null hypothesis, concluding that the observed effect is statistically significant.
T-Distribution
The t-distribution, also known as Student's t-distribution, is a probability distribution that is symmetric and bell-shaped, like the normal distribution, but has heavier tails. It arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown.

The t-distribution is wider than the normal distribution, reflecting the increased variability that arises from using a sample to estimate population parameters. As sample size increases, the t-distribution approaches the normal distribution. When performing a hypothesis test, the calculated t-value is compared to critical values from the t-distribution based on the degrees of freedom (df) from the sample data. The degrees of freedom for a t-test are typically the sample size minus one. In the exercise, the t-distribution provides a reference to evaluate how extreme the computed t-value is, which in turn helps determine the P-value and whether the null hypothesis can be rejected or not.

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Most popular questions from this chapter

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]: e953-e958) studied independent random samples of 764 Dutch boys and 748 Dutch girls ages 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. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls. Scenario 2: The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is \(\$ 48,722\). A random sample of 50 accounting graduates at a large university resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722\). Scenario 3: 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. You would like 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.

The paper referenced in the Preview Example of this chapter ("Mood Food: Chocolate and Depressive Symptoms in a Cross-Sectional Analysis," Archives of Internal Medicine [2010]: \(699-703\) ) describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on their test score. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings per month of chocolate for the sample of people that screened positive for depression was 8.39 , and the sample standard deviation was \(14.83 .\) For the sample of people who did not screen positive for depression, the mean was \(5.39,\) and the standard deviation was \(8.76 .\) The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants included 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is higher than the mean number of chocolate servings per month for people who would not screen positive.

The paper "Ladies First?" A Field Study of Discrimination in Coffee Shops" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05

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Babies born extremely prematurely run the risk of various neurological problems and tend to have lower IQ and verbal ability scores than babies who are not premature. The article "Premature Babies May Recover Intelligence, Study Says" (San Luis Obispo Tribune, February 12,2003 ) summarized medical research that suggests that the deficit observed at an early age may decrease as children age. Children who were born prematurely were given a test of verbal ability at age 3 and again at age 8 . The test is scaled so that a score of 100 would be average for normal-birth-weight children. Data that are consistent with summary quantities given in the paper for 50 children who were born prematurely were used to generate the accompanying Minitab output, where Age 3 represents the verbal ability score at age 3 and Age8 represents the verbal ability score at age \(8 .\) Use the information in the Minitab output to construct and interpret a \(95 \%\) confidence interval for the change in mean verbal ability score from age 3 to age 8 . You can assume that it is reasonable to regard the sample of 50 children as a random sample from the population of all children born prematurely. Paired T-Test and Cl: Age8, Age3 Paired \(\mathrm{T}\) for Age8 - Age3 \(\begin{array}{lrrrr} & \mathrm{N} & \text { Mean } & \text { StDev } & \text { Se Mean } \\ \text { Age8 } & 50 & 97.21 & 16.97 & 2.40 \\ \text { Age3 } & 50 & 87.30 & 13.84 & 1.96 \\ \text { Difference } & 50 & 9.91 & 22.11 & 3.13\end{array}\)

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