Not changed. A comparison was made between the...

not changed. A comparison was made between the patientreported favorable outcome and change in the MRI assessment of extent of nerve root compression (1 yr. vs. baseline). The results were as follows: Among 224 patients with a favorable outcome, nerve root compression had disappeared completely at 1 year for 169 patients, was present at 1 year but reduced from baseline for 39 patients, and was unchanged or increased from baseline for 16 patients. Among 43 patients with an unfavorable outcome, nerve root compression had disappeared completely at 1 year for 29 patients, was present at 1 year but reduced from baseline for 6 patients, and was unchanged or increased from baseline for 8 patients. Perform the test in Problem 10.56 , and report a two-sided $p$ -value.

not changed. A comparison was made between the patientreported favorable outcome and change in the MRI assessment of extent of nerve root compression (1 yr. vs. baseline). The results were as follows:

Among 224 patients with a favorable outcome, nerve root compression had disappeared completely at 1 year for 169 patients, was present at 1 year but reduced from baseline for 39 patients, and was unchanged or increased from baseline for 16 patients. Among 43 patients with an unfavorable outcome, nerve root compression had disappeared completely at 1 year for 29 patients, was present at 1 year but reduced from baseline for 6 patients, and was unchanged or increased from baseline for 8 patients.
Perform the test in Problem 10.56 , and report a two-sided $p$ -value.

Key Concepts

Contingency Table Analysis

In many studies involving categorical outcomes, data are organized into a contingency table. This table displays the frequency distribution of different outcomes in relation to categories, allowing researchers to visually and statistically assess associations between variables.

Hypothesis Testing

Hypothesis testing is the process used to evaluate assumptions about a population parameter, typically by comparing observed results with what would be expected under a null hypothesis. It involves formulating a null hypothesis, calculating a test statistic, and determining the p?value to decide whether to reject the null hypothesis.

Chi-Square Test of Independence

The chi-square test of independence is a statistical method used to determine whether there is a significant association between two categorical variables. By comparing observed frequencies with expected frequencies under the assumption of independence, this test helps to reveal whether any differences in the distribution of outcomes are due to chance or indicate a real relationship.

p-value

The p-value is the probability of observing an effect as extreme or more extreme than what was actually observed, assuming that the null hypothesis is true. In the context of a two-sided test, it accounts for deviations in both directions, and a small p-value indicates that the observed association is unlikely to be due to random variation.

Transcript

00:01 In problem 25, we have been given that three methods of reducing blood pressure were tested.

00:10 These methods are biofeedback, home relaxation, and a combination of biofeedback and relaxation.

00:19 And the sample sizes for this each group are given, 13, 14 and 15 respectively.

00:26 So it was found that there were no differences among the three groups.

00:30 And hence, a second test was done and there were significant differences between the groups in terms of their systolic measure and diastolic measure as given in this section with a very small p -values.

00:48 So in the first part of the question, we're going to verify that in the first test a, the the degrees of freedom for the methods equals 2 and the degrees of freedom for the error equals 39.

01:06 So how do we tell that this is true that the degrees of freedom for the method are 2? so the degrees of freedom for the factor, which in this case is the method, are all given by the formula c minus 1, where c stands for the number of factors.

01:26 And in this case we have three factors.

01:30 So c minus 1 will be 3 minus 1, which equals 2.

01:39 So that verifies that the degrees of freedom for the method are correct.

01:44 Next, we're going to look at the degrees of freedom for the error.

01:48 So the error degrees of freedom are given by the formula, n, which is the sample size, minus c, which is.

01:56 Is the number of factors.

01:58 And the total sample size in our case is 13 plus 14 plus 15.

02:08 And that gives us a total of 42.

02:11 So 42 minus 3, which is c, gives us 39, which verifies that the degrees of freedom for error are correct...