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91Ó°ÊÓ

A hospital readmission is an episode when a patient who has been discharged from a hospital is readmitted again within a certain period. Nationally the readmission rate for patients with pneumonia is \(17 \% .\) A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage. They found 11 patients out of 70 treated for pneumonia in a two-month period were readmitted. a. What is \(\hat{p}\), the sample proportion of readmission? b. Write the null and alternative hypotheses. c. Find the value of the test statistic and explain it in context. d. The p-value associated with this test statistic is \(0.39 .\) Explain the meaning of the \(\mathrm{p}\) -value in this context. Based on this result, does the \(\mathrm{p}\) -value indicate the null hypothesis should be doubted?

Short Answer

Expert verified
The p-value of 0.39 implies that the proportion of readmission as low (or even lower) than our calculated sample proportion would occur almost 39% of the time if the null hypothesis was true. Thus, there's not enough evidence to reject the null hypothesis.

Step by step solution

01

Finding the sample proportion (\(\hat{p}\))

We can calculate the sample proportion \(\hat{p}\) as the number of 'successes' divided by the total number of trials. Here, the successes are the patients readmitted i.e., 11 and the total number of trials are the total patients treated i.e., 70. Hence, \(\hat{p}\) = 11/70 = 0.157.
02

Formulating the null and alternative hypotheses

The null hypothesis (\(H_0\)) is that the sample proportion is equal to or more than the national readmission rate, and the alternative hypothesis (\(H_1\)) is that the sample proportion is less than the national rate. Hence, \(H_0: p \geq 0.17\), \(H_1: p < 0.17\).
03

Finding the value of the test statistic

We use the formula \((\hat{p} - p_{0}) / \sqrt{(p_{0}(1-p_{0})/n}\) to calculate the value of the test statistic. Here, \(p_0\) is the national readmission rate 0.17, \(\hat{p}\) is our calculated sample proportion 0.157 and \(n\) is the total number of patients treated i.e., 70. Plugging these values into the formula, we get a test statistic of -0.828.
04

Interpreting the test statistic in context

The negative value of the test statistic (-0.828) indicates that our observed sample proportion (the hospital’s readmission rate) is less than the national average readmission rate. However, it doesn't imply statistical significance. We must look at the p-value for that.
05

Interpreting the p-value in context and drawing conclusions about the hypotheses

A p-value of 0.39 is higher than the typical threshold of 0.05,Meaning the observed data was not extreme enough to reject the null hypothesis. This suggests that a proportion as low (or lower) than our sample proportion (\(\hat{p}\)= 0.157) would occur about 39% of the time under the null hypothesis. Therefore, the null hypothesis should not be doubted.

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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 structured process used in statistics to determine whether there's enough evidence in a sample of data to infer that a certain condition holds for the entire population. In the context of hospital readmission rates, hypothesis testing helps in ascertaining if the observed sample proportion of patients who were readmitted is significantly different from the national readmission rate.

Let's consider the scenario where a hospital wants to know if their readmission rate for pneumonia is lower than the national rate which is 17%. The first step is to establish null (\(H_0\text{): the hospital's rate is at least 17%}\) and alternative (\(H_1\text{): the hospital's rate is less than 17%}\) hypotheses. The test will determine if the sample data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis or not.
Sample Proportion
The sample proportion, denoted by \(\hat{p}\), is the ratio of the number of occurrences (or 'successes') to the total number of observations in the sample. In our case, the occurrences are the patients readmitted to the hospital after discharge for pneumonia.

To calculate the sample proportion, you divide the number of readmitted patients (11) by the total number treated for pneumonia (70), resulting in \(\hat{p} = 0.157\). This value represents the observed proportion of readmissions in the hospital's sample and acts as the empirical basis to test against the national average.
P-Value Interpretation
The p-value is a crucial concept in hypothesis testing. It serves as a bridge between the sample data and the hypotheses being tested. The p-value quantifies the probability that the observed data, or something more extreme, would occur under the assumption that the null hypothesis \(H_0\) is true.

In our hospital readmission example, a p-value of 0.39 means that if the true rate of readmission were actually 17% (or higher), as the null hypothesis states, there would be a 39% chance to observe a sample proportion as extreme as 0.157 (or more extreme), purely due to random variation in the sample. Since this p-value is not lower than the commonly used significance level of 0.05, we do not have enough evidence to reject the null hypothesis. Thus, it suggests that the sample does not provide strong enough evidence to conclude that the hospital's readmission rate is less than the national rate.
Test Statistic
The test statistic is a standardized value used in hypothesis testing to measure how far the sample proportion deviates from the null hypothesis. It's calculated using a specific formula that depends on the type of test being performed.

In our example, the test statistic is computed by the formula \[\frac{(\hat{p} - p_0)}{\sqrt{(p_0(1-p_0)/n)}}\] where \(p_0\) is the national readmission rate, \(\hat{p}\) is the sample proportion, and \(n\) is the sample size. The calculated value of -0.828 indicates that the observed proportion is less than the national average. However, the magnitude of this statistic alone does not tell us whether this difference is meaningful in a statistical sense. We rely on the p-value to interpret this test statistic within the context of hypothesis testing.

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Most popular questions from this chapter

According to the Bureau of Labor Statistics, \(10.1 \%\) of Americans are self- employed. A researcher wants to determine if the self-employment rate in a certain area is different. She takes a random sample of 500 working residents from the area and finds that 62 are self-employed. a. Test the hypothesis that the proportion of self-employed workers in this area is different from \(10.1 \%\). Use a \(0.05\) significance level. b. After conducting the hypothesis test, a further question one might ask, "What proportion of workers in this area are self-employed?" Use the sample data to find a \(95 \%\) confidence interval for the proportion of workers who are self-employed in the area from which the sample was drawn. How does this confidence interval support the hypothesis test conclusion?

A psychologist is interested in testing whether offering students a financial incentive improves their video-game-playing skills. She collects data and performs a hypothesis test to test whether the probability of getting to the highest level of a video game is greater with a financial incentive than without. Her null hypothesis is that the probability of getting to this level is the same with or without a financial incentive. The alternative is that this probability is greater. She gets a p-value from her hypothesis test of \(0.003 .\) Which of the following is the best interpretation of the p-value? i. The p-value is the probability that financial incentives are not effective in this context. ii. The p-value is the probability of getting exactly the result obtained, assuming that financial incentives are not effective in this context. iii. The p-value is the probability of getting a result as extreme as or more extreme than the one obtained, assuming that financial incentives are not effective in this context. iv. The p-value is the probability of getting exactly the result obtained, assuming that financial incentives are effective in this context. \(\mathrm{v}\). The p-value is the probability of getting a result as extreme as or more extreme than the one obtained, assuming that financial incentives are effective in this context.

Suppose we are testing people to see whether the rate of use of seat belts has changed from a previous value of \(88 \%\). Suppose that in our random sample of 500 people we see that 450 have the seat belt fastened. Which of the following figures has the correct p-value for testing the hypothesis that the proportion who use seat belts has changed? Explain your choice.

A magazine advertisement claims that wearing a magnetized bracelet will reduce arthritis pain in those who suffer from arthritis. A medical researcher tests this claim with 233 arthritis sufferers randomly assigned either to wear a magnetized bracelet or to wear a placebo bracelet. The researcher records the proportion of each group who report relief from arthritis pain after 6 weeks. After analyzing the data, he fails to reject the null hypothesis. Which of the following are valid interpretations of his findings? There may be more than one correct answer. a. The magnetized bracelets are not effective at reducing arthritis pain. b. There's insufficient evidence that the magnetized bracelets are effective at reducing arthritis pain. c. The magnetized bracelets had exactly the same effect as the placebo in reducing arthritis pain. d. There were no statistically significant differences between the magnetized bracelets and the placebos in reducing arthritis pain.

In 2015 a Gallup poll reported that \(52 \%\) of Americans were satisfied with the quality of the environment. In 2018 , a survey of 1024 Americans found that 461 were satisfied with the quality of the environment. Does this survey provide evidence that satisfaction with the quality of the environment among Americans has decreased? Use a \(0.05\) significance level.

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