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Car Window Skin Cancer? A new study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{43}\) The study reviewed the records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004 . Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

Short Answer

Expert verified
Although the data shows a higher prevalence of melanomas on the left side, it does not directly establish a cause-effect relationship with exposure to sunlight through car windows. The confidence interval suggests that melanomas are indeed more likely to occur on the left side, and the null hypothesis can be rejected based on the low p-value. However, further studies would be needed to confidently confirm the hypothesis that skin cancers are more prevalent on the left side due to the sunlight coming in through car windows.

Step by step solution

01

Determine the study type

The type of study is inferred based on its design. In this case, the investigators reviewed existing records, therefore it is an observational study.
02

Calculate the proportion

The proportion of patients with melanoma that had the cancer on the left side can be calculated as 31 (number of patients where the cancer occurred on the left side) divided by 42 (total number of patients with melanoma), which is approximately 0.738.
03

Interpret the confidence interval

A 95% confidence interval of 0.579 to 0.861 means that if the study were repeated multiple times, in 95% of those replications the true population proportion would fall between these two values.
04

State the null and alternative hypotheses

The null hypothesis (H0) would be: Melanomas are equally likely to occur on the left side and the right side. The alternative hypothesis (Ha) would be: Melanomas are more likely to occur on the left side than on the right side.
05

Determine the type of test

Since the question of interest is whether melanomas are more likely to occur on the left side, a one-tailed test would be appropriate.
06

Predict the test results based on the confidence interval

The lower limit of the confidence interval (0.579) is greater than 0.5. In a two-tail test, this would suggest a rejection of the null hypothesis at the 0.05 significance level. Therefore, it can be predicted that it would be rejected in a one-tail test at the same significance level.
07

Conclusion based on the p-value

A p-value of 0.003 indicates strong evidence against the null hypothesis, meaning there's a very small probability that such an extreme statistic could happen by chance given that the null hypothesis is true. Thus, the conclusion would be that melanomas are more likely to occur on the left side.
08

Evaluate the implications of the study

While the data shows a higher prevalence of melanomas on the left side, it does not directly establish a cause-effect relationship with exposure to sunlight through car windows. Further studies would be needed to confidently confirm this hypothesis.

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Key Concepts

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

Observational Study
An observational study is a type of research where investigators observe subjects and measure variables of interest without assigning treatments to the subjects. In the context of skin cancer statistics and the Car Window Skin Cancer study, the researchers reviewed patient records to collect data, rather than manipulating any variables or conditions. This approach allows for the collection of real-world data, but it comes with limitations.

One major limitation of observational studies is their inability to prove causation. While an association can be identified, such as the higher incidence of melanoma on the left side of patients, additional investigative methods are required to establish a direct cause-and-effect relationship. Observational studies are excellent for generating hypotheses but require experimental or more advanced statistical methods to further test these hypotheses.

Improving an observational study involves methods like adjusting for confounding variables, using larger sample sizes, and longitudinal tracking to better understand the relationships between variables.
Confidence Interval
A confidence interval (CI) provides a range of values, derived from the data, that is likely to contain the population parameter with a specified level of confidence. In our exercise, a 95% confidence interval from 0.579 to 0.861 for melanoma occurrence on the left side suggests that we can be 95% confident that the true proportion of left-side melanomas in the population falls within this range.

The interpretation is fundamental: The CI does not mean that 95% of individual cases will fall within the range. Instead, it communicates that if we were to conduct the same study many times, 95% of the calculated intervals would capture the true population parameter. To increase the precision of a CI, researchers can either increase the sample size or choose a larger confidence level. However, a larger confidence level would also yield a wider interval, possibly making it less useful in certain decision-making contexts.

Understanding CIs is critical as they provide insight into the reliability and stability of estimates, leading to better-informed decisions in both science and policy.
Hypothesis Testing
Hypothesis testing is a statistical method that allows researchers to make inferences about a population based on sample data. The null hypothesis (H0) represents a default statement that there is no effect or no difference, while the alternative hypothesis (Ha) represents what we are trying to provide evidence for.

In the skin cancer study, the test was whether melanomas are more likely to occur on the left side. The null hypothesis in this scenario is that the occurrence of melanoma is equal on both sides, and the alternative hypothesis is that there is a higher occurrence on the left side. The p-value, which in this case is 0.003, indicates the probability of observing the data if the null hypothesis were true; a small p-value suggests that the observed data is quite unlikely under H0, thus providing evidence to reject H0 in favor of Ha.

One critical aspect of hypothesis testing is determining whether to use a one-tailed or two-tailed test. This decision hinges on whether the research question is looking for a specific direction of association or any significant association at all. In the provided case, a one-tailed test is suitable since the hypothesis specifically concerns a higher incidence on one side. Hypothesis testing allows researchers to make probabilistic conclusions about whether their data provides enough evidence to support a given hypothesis, therefore playing a crucial role in many fields of scientific inquiry.

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Most popular questions from this chapter

In Exercises 4.150 to \(4.152,\) a confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A 99\% confidence interval for \(\mu: 134\) to 161 (a) \(H_{0}: \mu=100\) vs \(H_{a}: \mu \neq 100\) (b) \(H_{0}: \mu=150\) vs \(H_{a}: \mu \neq 150\) (c) \(H_{0}: \mu=200\) vs \(H_{a}: \mu \neq 200\)

In Exercises 4.112 to \(4.116,\) the null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: \rho=0\) vs \(H_{a}: \rho \neq 0\) Sample: \(r=-0.29, n=50\) Randomization statistic \(=r\)

Classroom Games Two professors \(^{18}\) at the University of Arizona were interested in whether having students actually play a game would help them analyze theoretical properties of the game. The professors performed an experiment in which students played one of two games before coming to a class where both games were discussed. Students were randomly assigned to which of the two games they played, which we'll call Game 1 and Game \(2 .\) On a later exam, students were asked to solve problems involving both games, with Question 1 referring to Game 1 and Question 2 referring to Game 2 . When comparing the performance of the two groups on the exam question related to Game 1 , they suspected that the mean for students who had played Game 1 ( \(\mu_{1}\) ) would be higher than the mean for the other students \(\mu_{2},\) so they considered the hypotheses \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) (a) The paper states: "test of difference in means results in a p-value of \(0.7619 . "\) Do you think this provides sufficient evidence to conclude that playing Game 1 helped student performance on that exam question? Explain. (b) If they were to repeat this experiment 1000 times, and there really is no effect from playing the game, roughly how many times would you expect the results to be as extreme as those observed in the actual study? (c) When testing a difference in mean performance between the two groups on exam Question 2 related to Game 2 (so now the alternative is reversed to be \(H_{a}: \mu_{1}<\mu_{2}\) where \(\mu_{1}\) and \(\mu_{2}\) represent the mean on Question 2 for the respective groups), they computed a p-value of \(0.5490 .\) Explain what it means (in the context of this problem) for both p-values to be greater than \(0.5 .\)

Red Wine and Weight Loss Resveratrol, an ingredient in red wine and grapes, has been shown to promote weight loss in rodents. A recent study \(^{19}\) investigates whether the same phenomenon holds true in primates. The grey mouse lemur, a primate, demonstrates seasonal spontaneous obesity in preparation for winter, doubling its body mass. A sample of six lemurs had their resting metabolic rate, body mass gain, food intake, and locomotor activity measured for one week prior to resveratrol supplementation (to serve as a baseline) and then the four indicators were measured again after treatment with a resveratrol supplement for four weeks. Some p-values for tests comparing the mean differences in these variables (before vs after treatment) are given below. In parts (a) to (d), state the conclusion of the test using a \(5 \%\) significance level, and interpret the conclusion in context. (a) In a test to see if mean resting metabolic rate is higher after treatment, \(p=0.013\). (b) In a test to see if mean body mass gain is lower after treatment, \(p=0.007\) (c) In a test to see if mean food intake is affected by the treatment, \(p=0.035\). (d) In a test to see if mean locomotor activity is affected by the treatment, \(p=0.980\) (e) In which test is the strongest evidence found? The weakest? (f) How do your answers to parts (a) to (d) change if the researchers make their conclusions using a stricter \(1 \%\) significance level? (g) For each p-value, give an informal conclusion in the context of the problem describing the level of evidence for the result. (h) The sample only included six lemurs. Do you think that we can generalize to the population of all lemurs that body mass gain is lower on average after four weeks of a resveratrol supplement? Why or why not?

Arsenic in Chicken Data 4.5 on page 228 introduces a situation in which a restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than 80 ppb, so we are testing \(H_{0}: \mu=80\) vs \(H_{a}: \mu>80\), where \(\mu\) represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic so samples are often small. Suppose \(n=6\) chickens from one supplier are tested, and the levels of arsenic (in ppb) are: \(68, \quad 75\) 81, \(\quad 93\) 134 (a) What is the sample mean for the data? (b) Translate the original sample data by the appropriate amount to create a new dataset in which the null hypothesis is true. How do the sample size and standard deviation of this new dataset compare to the sample size and standard deviation of the original dataset? (c) Write the six new data values from part (b) on six cards. Sample from these cards with replacement to generate one randomization sample. (Select a card at random, record the value, put it back, select another at random, until you have a sample of size \(6,\) to match the original sample size.) List the values in the sample and give the sample mean. (d) Generate 9 more simulated samples, for a total of 10 samples for a randomization distribution. Give the sample mean in each case and create a small dotplot. Use an arrow to locate the original sample mean on your dotplot.

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