In Exercises 1–10, use a 0.05 significance level with the indi...

Question

In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

World Series The last 114 baseball World Series ended with 66 wins by American League teams and 48 wins by National League teams. Use the sign test to test the claim that in each World Series, the American League team has a 0.5 probability of winning.

Accepted Answer

Hello everyone. Let's take a look at this question together. A new teaching method is tried in 102 classes. It leads to improved scores in 60 classes and no improvement in 42 classes. At the 0.05 significance level, use the sign test to test the claim that improvements and no improvements are equally likely. So the first step in solving this problem is to state our hypotheses, where our null hypothesis is that improvements and no improvements are equally likely. With a P value equal to 0.5, and our alternative hypothesis is that improvements and no improvements are not equally likely as our P-value is not equal to 0.5. So then the next step is to identify the sample size and the number of successes. Where the sample size or the total number of classes is represented by N, which is equal to 60 + 42, so our total number of classes, which is our sample size, is 102. And then the number of improvements, which is the number of successes, is represented by X and is equal to 60, and using our data, we can calculate the test statistic using the binomial distribution where under the null hypothesis, the number of improvements follows X is distributed binominally with N equals 102 and. equals 0.5. And then for a large n value, we use the normal approximation, which is the mean is equal to mu, which equals n multiplied by P. So plugging in the values we have 102 multiplied by 0.5, so the mean mu is equal to 51, and then the standard deviation is calculated as sigma equals the square root. Of N multiplied by P which is multiplied by 1 minus P in parentheses, and plugging in the values into that formula, our standard deviation is approximately 5.05. And then we can calculate the Z score with continuity correction where using continuity correction X equals 60 becomes 59.5. And then our Z score is equal to 59.5 minus 51 divided by 5.05, so the Z score is approximately 1.68. And then we find the P value, which for a two-tailed test, we are finding the probability that the absolute value of Z is greater than 1.68. And from the standard normal distribution table, the probability of Z being less than 1.68 is approximately 0.9535. Therefore, the probability of Z being greater than 1.68 is approximately 1 minus 0.9535, so it is approximately 0.0465. So then our P value is equal to 2 multiplied by 0.0465, which is 0.093. And comparing the P value to the significance level, since 0.093 is greater than 0.05, we fail to reject the null hypothesis. So in conclusion, at the 0.05 significance level, there is insufficient evidence to support the claim that improvements and no improvements are not equally likely. So this makes answer choice be the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Sign Test

Null Hypothesis and Alternative Hypothesis

Significance Level

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