In Exercises 5–8, use (a) randomization and (b) bootstrapping ...

Question

In Exercises 5–8, use (a) randomization and (b) bootstrapping for the indicated exercise from Section 9-1. Compare the results to those obtained in the original exercise.

Exercise 8 in Section 9-1 “Tennis Challenges”

Accepted Answer

All right, hello, everyone. So this question says researchers are comparing the rates at which two different hospitals successfully treat a certain infection. Hospital X treated 950 patients and had 320 successful recoveries. Hospital Y treated 780 patients and had 255 successful recoveries. At the 0.05 significance level, test whether the success rates are the same for both hospitals using a randomization and B bootstrapping. So before we get into either of the tests, what information are we actually given? Well, we're given the success rates for both hospitals and we're also able to find the observed difference between them. So for Hospital X. He had Of X which corresponds to the rate of successful treatments would be equal to. 320 Divided by 950. Which gives you approximately 0.337. For hospital Y he had. Would be equal to. 255 divided by 780. And this equals approximately 0.327. So now the observed difference between them. Is simply the difference between Phat of X and Phat of Y. Which is approximately 0.010. So now let's begin with the randomization or permutation test for Part A. Now, the goal of the randomization test is to test if the observed difference in success rates could arise by random chance. So first, what are the hypotheses, right? The null hypothesis states that. P X or Pha of X is equal to Pha of Y. On the other hand, the alternative hypothesis. States that Pha of X is not equal to Pha Y, meaning that they have different success rates. So, because we're assuming that. That both Phat values are equal under the null hypothesis, we want to find the pooled success rate, or just Phat. So the pooled success rate is equal to the sum of successful treatments in both hospitals divided by the total number of patients in both hospitals. So in this case, that's the. Phad is equal to the sum of 320 and 255. Divided by the sum of 950 and 780. And so this equals approximately 0.332. So now what do you do at this point? While you would simulate under the null hypothesis, that is, you would assign. 950 patients to hospital X, and 780 to hospital Y from the pooled data. For each simulation you would then calculate. The difference between Pha of X and P Y. From there, you would repeat many, many times, upwards of 10,000 iterations to build the null distribution. And then you calculate the p value. Now, the P value in this context is the proportion of simulated differences as or more extreme than the difference of 0.010. If the p-value is less than 0.05, at that point you can reject the null hypothesis. But what outcome do we expect given the information we have? Well, notice how the difference is quite small, right? It's 0.010, which means that the P value is likely to instead be greater than 0.05, which means That we would expect To fail to reject The no hypothesis. So now let's move on with bootstrapping. Now with bootstrapping, the goal here is to estimate the uncertainty. Associated with the true difference that is PX subtracted by PY. So, the first step for bootstrapping for the 95% confidence interval is to resample with replacement. So for Hospital X, you would draw 950 patients with replacement from the original data, and you would do the same thing for hospital Y with 780 patients from the original data. And then For each bootstrap sample, you would calculate PX subtracted by P Y. And then you would repeat this process 10,000 times to get a distribution of differences before constructing the 95% confidence interval. Now, for the 95% confidence interval, using the percentile method, the bootstrap distribution would use the 2.5. And the 97.5. Percentiles So what precisely is the expected outcome of this case? Well, the 95% confidence interval is likely going to include the number 0. Which would therefore support the null hypothesis. So what exactly have we gathered up to this point? Well, according to the randomization test, we failed to reject the null hypothesis because the P value would be greater than 0.05. And according to the bootstrap 95% confidence interval, because it's likely to include zero, that suggests no statistically significant difference. Which brings us now to our final answer. Which states that neither method shows a statistically significant difference in success rates. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

In Exercises 5–8, use (a) randomization and (b) bootstrapping for the indicated exercise from Section 9-1. Compare the results to those obtained in the original exercise.

Exercise 8 in Section 9-1 “Tennis Challenges”

Key Concepts

Randomization

Bootstrapping

Comparison of Results

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