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Treatment with statins can reduce the risk of a major cardiovascular event in people with specified risk factors. During about 2 years of follow-up in the JUPITER trial (P. M. Ridker et al. \([8]), 142\) of 8901 subjects treated with a statin had a major cardiovascular event. Suppose the expected 2-year risk of a major cardiovascular event in similar but untreated people is 0.028 . Test whether the 2 -year risk in those treated with a statin is significantly different from this expected risk at the \(5 \%\) level. Make sure to state your null and alternative hypotheses and specify your conclusion.

Short Answer

Expert verified
With the given data, the 2-year risk for those treated with a statin is significantly different from the expected risk of 0.028.

Step by step solution

01

Define Hypotheses

The null hypothesis (\(H_0\)) is that the 2-year risk of a major cardiovascular event for those treated with a statin is equal to the expected risk for untreated individuals, i.e., \(p = 0.028\). The alternative hypothesis (\(H_a\)) is that the 2-year risk is different from the expected risk, i.e., \(p eq 0.028\).
02

Calculate the Proportion of Events Among Treated

Determine the sample proportion of subjects treated with a statin who had a major cardiovascular event. Calculate it as \(\hat{p} = \frac{142}{8901} \approx 0.0159\).
03

Determine the Test Statistic

For a proportion, the test statistic \(z\) is calculated using \(z = \frac{\hat{p} - 0.028}{\sqrt{\frac{0.028 \times (1 - 0.028)}{8901}}}\). Compute this step to get the \(z\)-value.
04

Calculate the Standard Error

Calculate the standard error (SE) as follows: \(SE = \sqrt{\frac{0.028 \times (1 - 0.028)}{8901}} = \sqrt{\frac{0.028 \times 0.972}{8901}}\). Compute SE for further calculations.
05

Compute the Z-Value

Substitute the values into the formula from Step 3 to get the \(z\)-value. \(z = \frac{0.0159 - 0.028}{SE}\). Compute \(z\)-value to assess significance.
06

Find the Critical Z-Value and Make a Decision

For a two-tailed test at the 5% significance level, the critical \(z\)-values are approximately \( \pm 1.96\). Compare the computed \(z\)-value with the critical value: if the computed \(z\) is beyond \(\pm 1.96\), reject the null hypothesis.
07

Conclusion

If the computed \(z\)-value lies outside the range of \(-1.96\) to \(1.96\), there is significant evidence to reject the null hypothesis. Otherwise, there is not enough evidence to conclude the risk is different.

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Key Concepts

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

Null Hypothesis
The concept of a null hypothesis is fundamental in hypothesis testing. In our scenario, the null hypothesis (\(H_0\)) posits that there is no difference in the event rate between those who are taking statins and the expected rate in untreated individuals. To put it simply, it maintains that the treatment has no effect.

In statistical terms, it's often formulated as an equality. Here, we stated it as \(p = 0.028\), indicating that the probability of an event (a cardiovascular incident) is the same for both groups. This assumption is critical because it forms the baseline against which we compare our observed data.

The null hypothesis serves several purposes:
  • It provides a claim to test against.
  • It often represents the current accepted state before any change (i.e., no difference).
  • It's presumed true until evidence suggests otherwise.
In practice, rejecting the null hypothesis suggests that the evidence is strong enough to support a change or difference indicated by an alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) is the counterpart to the null hypothesis and proposes that there is a statistically significant difference to consider. In the context of our statin study, it's the hypothesis suggesting a difference in cardiovascular event rates due to the treatment.

To articulate this, the alternative hypothesis is often expressed as not equal (\(p eq 0.028\)) indicating any difference—higher or lower—in the risk when treated with statins. This formulation in hypothesis testing is crucial for considering:
  • Changes that could lead to rejecting the null hypothesis.
  • The positive or negative effects resulting from treatments or interventions.
  • Understanding the directional nature of the research question being tested.
Recognizing this hypothesis allows researchers to be ready for any outcome and aligns closely with what you're aiming to investigate: the effectiveness or impact of a treatment.
Standard Error
Standard error is a statistical measure that provides insight into the amount of variability in the estimate of a population parameter. It plays a crucial role in hypothesis testing, especially when comparing an observed value against a hypothesized value.

In our case, the standard error (SE) related to the proportion of cardiovascular events in the treated group was calculated using:\[SE = \sqrt{\frac{0.028 \times (1 - 0.028)}{8901}} = \sqrt{\frac{0.028 \times 0.972}{8901}}\]This measure tells us how much the sample proportion (\(\hat{p}\)) might vary around the true population proportion. A smaller SE indicates a more precise estimate of the population proportion, strengthening the reliability of conclusions drawn from the hypothesis test.

Key functions of standard error include:
  • Determining the test statistic (like the z-value in our example).
  • Serving as a critical input for forming confidence intervals.
  • Helping assess the reliability of predictions in statistical conclusions.
Ultimately, understanding SE is essential for interpreting how sample data extrapolates to larger populations and making informed, statistically sound decisions.

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Most popular questions from this chapter

A clinical epidemiologic study was conducted to determine the long-term health effects of workplace exposure to the process of manufacturing the herbicide ( 2,4,5 trichlorophenoxy) acetic acid \((2,4,5-\mathrm{T}),\) which contains the contaminant dioxin [7]. This study was conducted among active and retired workers of a Nitro, West Virginia, plant who were exposed to the \(2,4,5-T\) process between 1948 and 1969 . It is well known that workers exposed to 2,4,5 -T have high rates of chloracne (a generalized acneiform eruption). Less well known are other potential effects of \(2,4,5-T\) exposure. One of the variables studied was pulmonary function. Suppose the researchers expect from general population estimates that \(5 \%\) of workers have an abnormal forced expiratory volume (FEV); defined as less than \(80 \%\) of predicted, based on their age and height. They found that 32 of 203 men who were exposed to \(2,4,5-T\) while working at the plant had an abnormal FEV. What hypothesis test can be used to test the hypothesis that the percentage of abnormal FEV values among exposed men differs from the general-population estimates?

Another issue investigated in this study was the case-fatality rate (number of patients who died from neuroblastoma / number of cases identified by the screening program). Suppose the case-fatality rate from neuroblastoma is usually \(1 / 6 .\) Furthermore, 17 fatal cases occurred among the 204 cases identified in the screening program. Test whether the case-fatality rate under the screening program is different from the usual case-fatality rate. Provide a two-tailed \(p\) -value.

Use a computer program to compute the probability that a \(t\) distribution with 36 df exceeds 2.5.

Osteoporosis is an important cause of morbidity in middle-aged and elderly women. Several drugs are currently used to prevent fractures in postmenopausal women. Suppose the incidence rate of fractures over a 4 -year period is known to be \(5 \%\) among untreated postmenopausal women with no previous fractures. A pilot study conducted among 100 women without previous fractures aims to determine whether a new drug can prevent fractures. It is found that two of the women have developed fractures over a 4-year period. Suppose the new drug is hypothesized to yield a fracture rate of \(2.5 \%\) over a 4 -year period. How many subjects need to be studied to have an \(80 \%\) chance of detecting a significant difference between the incidence rate of fractures in treated women and the incidence rate of fractures in untreated women (assumed to be \(5 \%\) from Problem 7.105 )?

A study was performed among patients with glaucoma, an important eye disease usually manifested by high intraocular pressure (IOP); left untreated, glaucoma can lead to blindness. The patients were currently on two medications (A and B) to be taken together for this condition. The purpose of this study was to determine whether the patients could drop medications \(A\) and \(B\) and be switched to a third medication (medication C) without much change in their IOP. Ten patients were enrolled in the study. They received medications \(A+B\) for 60 days and had their IOP measured at the end of the 60-day period (referred to as \(\mathrm{IOP}_{\mathrm{A}+\mathrm{B}}\) ). They were then taken off medications \(A\) and \(B\) and switched to medication \(C\), which they took for an additional 60 days. IOP was measured a second time at the end of the 60 -day period while the patient was on medication \(\mathrm{C}\) (referred to as \(\mathrm{IOP}_{\mathrm{C}}\) ). The results were as shown in Table 7.12. What procedure can be used to test the hypothesis that there has been no mean difference in IOP after 60 days between the two drug regimens?

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