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Chapter 11: Problem 38

Preventing strokes Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anticlotting drug named dipyridamole would be more effective for patients who had already had a stroke. Here are the data on strokes during the two years of the study: \(^{20}\). $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Number of patients } & \text { Number of strokes } \\ 1 & \text { Placebo } & 1649 & 250 \\ 2 & \text { Aspirin } & 1649 & 206 \\ 3 & \text { Dipyridamole } & 1654 & 211 \\ 4 & \text { Both } & 1650 & 157 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the stroke rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=p_{2}=p_{3}=p_{4}\) says about the incidence of strokes. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

Short Answer

Expert verified
The stroke rates differ among the treatments, specifically, the "Both" treatment is most effective.

Step by step solution

01

Create a Two-Way Table

We begin by summarizing the data in a two-way table with treatment groups as rows and outcomes (Number of Patients and Number of Strokes) as columns: | Group | Treatment | Number of Patients | Number of Strokes | |-------|----------------|-------------------|------------------| | 1 | Placebo | 1649 | 250 | | 2 | Aspirin | 1649 | 206 | | 3 | Dipyridamole | 1654 | 211 | | 4 | Both | 1650 | 157 |
02

Calculate Stroke Rates

Calculate the stroke rate for each treatment group by dividing the Number of Strokes by the Number of Patients:- Placebo: \( \frac{250}{1649} \approx 0.1517 \)- Aspirin: \( \frac{206}{1649} \approx 0.1249 \)- Dipyridamole: \( \frac{211}{1654} \approx 0.1276 \)- Both: \( \frac{157}{1650} \approx 0.0952 \)These rates show the proportion of patients who had strokes in each group.
03

State the Null Hypothesis

The null hypothesis, \( H_0: p_1 = p_2 = p_3 = p_4 \), states there is no difference in the stroke rates among the four treatment groups. In other words, the incidence of strokes is the same across all groups.
04

Conduct a Chi-Square Test

To determine if there is a significant difference in stroke rates, a chi-square test of independence can be performed. Calculate the expected number of strokes for each group assuming the null hypothesis is true, then use the formula for chi-square:\[\chi^2 = \sum \frac{(O - E)^2}{E}\]Where \(O\) is the observed frequency (strokes per group), and \(E\) is the expected frequency under the null hypothesis.
05

Compare Chi-Square Statistic to Critical Value

Determine the critical value of the chi-square distribution for \(\alpha = 0.05\) with 3 degrees of freedom (since there are 4 groups - 1) using a chi-square table. Compare the calculated chi-square statistic to this critical value to decide whether to reject \(H_0\).
06

Draw Conclusion

If the calculated chi-square statistic is greater than the critical value, reject the null hypothesis, indicating that there is a statistically significant difference in stroke rates among the treatment groups. If it's less, fail to reject \(H_0\).

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is key when analyzing experiments like the Stroke Prevention Study. The null hypothesis, represented as \( H_0: p_1 = p_2 = p_3 = p_4 \), suggests that there is no variation in the stroke rates among different treatment groups. This means that whether patients received a placebo, aspirin, dipyridamole, or both aspirin and dipyridamole, the incidence of strokes would remain consistent.
The premise is that all treatments would have an equal effect on patients, and any observed differences in stroke rates are merely due to random chance.

The null hypothesis serves as a baseline assumption. Researchers aim to test this hypothesis rigorously to determine if it can be rejected based on statistical evidence. In other words, by analyzing the data, they seek to find out if there is any meaningful difference between the effectiveness of the treatments that deviate from the expected equal outcomes.
Stroke Prevention Study
The Stroke Prevention Study is a practical investigation designed to evaluate if adding the drug dipyridamole to aspirin therapy can further reduce stroke occurrences.
Strokes are a serious health issue, often occurring due to blood clots in the brain, so researchers are constantly seeking new methods to prevent them.

This particular study involved four groups:
  • One received a placebo and served as a control group.
  • One received aspirin, a known anticlotting agent.
  • One received dipyridamole, another anticlotting agent.
  • The last group received both aspirin and dipyridamole.
Each group had a similar number of participants, and the study sought to compare the incidence of strokes across these different treatments over two years.

The goal was to identify which treatment, if any, offered superior stroke prevention. By carefully monitoring the number of strokes occurring in each group, researchers could analyze how effective each treatment was in preventing stroke events.
Significance Level
In statistical testing, the significance level, denoted by \( \alpha \), is a critical threshold used to determine the cutoff for rejecting the null hypothesis. In this study, a significance level of 0.05 was chosen, which is a common standard.
A significance level of 0.05 implies that there is a 5% risk of concluding that a difference exists when there is no actual difference.

This threshold helps to balance between being too lenient (risking false positives) and too strict (risking false negatives). In practice, it means that if the probability of observing the data, or something more extreme, is less than 5% under the assumption of the null hypothesis, then the result is considered statistically significant.

With a 0.05 significance level, researchers could confidently accept or reject the null hypothesis by comparing the calculated p-value from their tests against this threshold. If the p-value is less than 0.05, it indicates there is enough evidence to reject the null hypothesis, suggesting that the difference in treatments is statistically significant.
Two-Way Table
A two-way table is an efficient way to organize and display categorical data in studies. In the context of the Stroke Prevention Study, it helps researchers observe and compare stroke incidences among different treatment groups. The table outlines:
  • Four rows, each representing a different treatment group: placebo, aspirin, dipyridamole, and both drugs.
  • Columns show data for the number of patients and the number of strokes observed per treatment.
This visual arrangement simplifies the comparison process by easily showing the relationship between treatments and outcomes.

Comparing the numbers in this table allows researchers to quickly compute and analyze the stroke rates for each category. It provides the data needed to calculate the chi-square statistic鈥攁 measure used to assess whether observed differences among groups are statistically significant. Using this data, the chi-square test can support or refute the null hypothesis based on whether the actual frequencies deviate significantly from expected frequencies.

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Most popular questions from this chapter

Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.

No chi-square A school's principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night: $$ \begin{array}{lccccccc} \hline \text { Night: } & \text { Sunday } & \text { Monday } & \text { Tuesday } & \text { Wednesday } & \text { Thursday } & \text { Friday } & \text { Saturday } \\ \text { Average } & 130 & 108 & 115 & 104 & 99 & 37 & 62 \\ \text { time: } & & & & & & & \\ \hline \end{array} $$ Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

You may find the inference summary chart inside the back cover helpful. Inference recap \((8.1\) to 11.2\()\) In each of the following settings, state which inference procedure from Chapter \(8,9,10,\) or 11 you would use. Be specific. For example, you might say "two-sample \(z\) test for the difference between two proportions." You do not need to carry out any procedures. \(^{30}\) (a) Is there a relationship between attendance at religious services and alcohol consumption? A random sample of 1000 adults was asked whether they regularly attend religious services and whether they drink alcohol daily. (b) Separate random samples of 75 college students and 75 high school students were asked how much time, on average, they spend watching television each week. We want to estimate the difference in the average amount of \(\mathrm{TV}\) watched by high school and college students.

Seagulls by the seashore Do seagulls show a preference for where they land? To answer this question, biologists conducted a study in an enclosed outdoor space with a piece of shore whose area was made up of \(56 \%\) sand, \(29 \%\) mud, and \(15 \%\) rocks. The biologists chose 200 seagulls at random. Each seagull was released into the outdoor space on its own and observed until it landed somewhere on the piece of shore. In all, 128 seagulls landed on the sand, 61 landed in the mud, and 11 landed on the rocks. Do these data provide convincing evidence that seagulls show a preference for where they land?
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