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In the experiment described in the article "Study Points to Benefits of Knee Replacement Surgery Over Therapy Alone" (The New York Times, October 21,2015 ), adults who were considered candidates for knee replacement were followed for one year. Suppose that 200 patients were randomly assigned to one of two groups. One hundred were assigned to a group that had knee replacement surgery followed by therapy and the other half were assigned to a group that did not have surgery but did receive therapy. After one year, \(86 \%\) of the patients in the surgery group and \(68 \%\) of the patients in the therapy only group reported pain relief. Is there convincing evidence that the proportion experiencing pain relief is greater for the surgery treatment than for the therapy treatment? Use a significance level of 0.05

Short Answer

Expert verified
In conclusion, by performing a hypothesis test comparing the proportions of pain relief in the two groups with a significance level of 0.05, we obtained a test statistic, z 鈮 4.37, which resulted in a very small p-value. Since the p-value is smaller than the given significance level (伪 = 0.05), we can reject the null hypothesis and conclude that there is convincing evidence that the proportion of patients experiencing pain relief is greater for the surgery treatment group than for the therapy-only treatment group.

Step by step solution

01

State the Null and Alternative Hypotheses

We start by setting up our null and alternative hypotheses. The null hypothesis states that there is no difference in the proportion of pain relief between the two treatments, while the alternative hypothesis states that there is a difference in the proportion of pain relief, with the surgery treatment being more effective. H鈧: p鈧 = p鈧 H鈧: p鈧 > p鈧 Where p鈧 is the proportion of pain relief in the surgery group and p鈧 is the proportion of pain relief in the therapy-only group.
02

Calculate Observed Proportions and Test Statistic

Next, we need to calculate the observed proportions of pain relief for both groups. From the given data, we know that 86% of the surgery group and 68% of the therapy group reported pain relief. Observed p鈧 = 0.86 Observed p鈧 = 0.68 Now, we will calculate the test statistic using the following formula: z = \(\frac{(\text{observed p鈧亇 - \text{observed p鈧倉) - (\text{null p鈧亇 - \text{null p鈧倉)}{\sqrt{\frac{\text{null p} (1-\text{null p})}{n鈧亇 + \frac{\text{null p} (1-\text{null p})}{n鈧倉}}\) Where null_p鈧 and null_p鈧 are the assumed proportions under the null hypothesis (which are equal), null_p is the pooled sample proportion, and n鈧 and n鈧 are the sample sizes of each group.
03

Calculate Pooled Sample Proportion and Test Statistic

First, let's calculate the pooled sample proportion, null_p: null_p = \(\frac{0.86 \times 100 + 0.68 \times 100}{100 + 100} = \frac{154}{200} = 0.77\) Now we can calculate the test statistic, z: z = \(\frac{(0.86 - 0.68) - (0.77 - 0.77)}{\sqrt{\frac{0.77(1-0.77)}{100}+\frac{0.77(1-0.77)}{100}}}\) = \(\frac{0.18}{0.0412} \approx 4.37\)
04

Find the p-value

Now that we have our test statistic, we can find the p-value. Since this is a one-tailed test with an alternative hypothesis of p鈧 > p鈧, we will find the probability of getting a z-score greater than 4.37 in a standard normal distribution: p-value = P(Z > 4.37) A z-score of 4.37 is quite far out in the tail of the distribution, so the p-value will be very small.
05

Make a Decision

Finally, we compare our p-value to the given significance level (伪 = 0.05). Since our p-value is considerably smaller than 0.05, we can reject the null hypothesis and conclude that there is convincing evidence that the proportion of patients experiencing pain relief is greater for the surgery treatment group than for the therapy-only treatment group.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in statistics used during hypothesis testing. It is the claim that there is no effect or no difference, and it sets the stage for further statistical testing. In the context of our knee replacement study, the null hypothesis \( H_0 \) posits that the probability of experiencing pain relief from knee replacement surgery and therapy is the same as receiving only therapy.

Mathematically, it can be expressed as \( p_1 = p_2 \), where \( p_1 \) is the proportion of patients who experienced pain relief after surgery followed by therapy, and \( p_2 \) is the proportion of those who found relief with just therapy.

The null hypothesis serves as the "default" or "no effect" statement, meaning that any observed difference in the treatment effects is due to random chance instead of a true difference in the population. Rejecting the null would indicate evidence of a true difference, while failing to reject it suggests insufficient evidence to claim such a difference exists.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis and suggests that there is indeed an effect or a difference to be found. In our study on knee replacement and therapy, the alternative hypothesis \( H_1 \) asserts that the proportion of patients experiencing pain relief from surgery and therapy is greater than that from therapy alone.

Mathematically, this is expressed as \( p_1 > p_2 \). Here, \( p_1 \) and \( p_2 \) again represent the proportions of pain relief in the surgery and therapy-only groups, respectively.

The alternative hypothesis represents what the researcher aims to provide evidence for. In hypothesis testing, our goal is to determine if we have enough statistical support to reject the null hypothesis in favor of the alternative hypothesis, showing a substantive difference in outcomes between the two treatments.
Z-Test
A Z-test is a statistical method that helps in determining if there is a significant difference between two sample means. In the case of our knee replacement study, the Z-test checks whether the difference in pain relief proportions between the surgery and therapy groups is statistically significant.

The Z-test calculates a test statistic, known as the z-score, which measures how many standard deviations the observed data is from the expected data under the null hypothesis. To do this, we first calculate the pooled proportion and then use the formula: \[ z = \frac{(\text{observed } p_1 - \text{observed } p_2) - 0}{\sqrt{\frac{\text{null } p (1-\text{null } p)}{n_1} + \frac{\text{null } p (1-\text{null } p)}{n_2}}} \]

Here, the computed z-score was approximately 4.37, indicating the difference observed was substantial enough compared to what would be expected under the null hypothesis.
P-Value
The p-value is a key concept in hypothesis testing, giving the probability of observing test results at least as extreme as the results obtained, under the assumption that the null hypothesis is correct. A very low p-value suggests that the data are inconsistent with the null hypothesis and supports accepting the alternative hypothesis.

In the knee replacement study, the p-value must be compared to a significance level (\( \alpha \)) of 0.05. The z-score for this study was quite high (4.37), leading to a very small p-value, smaller than the significance threshold of 0.05.

With such a low p-value, we conclude with "statistical significance"鈥攖here is strong evidence against the null hypothesis. This means we can confidently state that the proportion of relief is greater in the surgery group compared to the therapy-only group.

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Most popular questions from this chapter

The report "Young People Living on the Edge" (Greenberg Quinlan Rosner Research, 2008) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (age 19 to 35 ), and the other sample consisted of 300 parents of young adults age 19 to \(35 .\) The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support in that situation. When asked about getting married, \(41 \%\) of the young adults said they thought parents would provide financial support and \(43 \%\) of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think their parents would provide financial support and the proportion of parents who say they would provide support are different.

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