A pediatrician claims that the mean birth weight of a single-b...

Question

A pediatrician claims that the mean birth weight of a single-birth baby is greater than the mean birth weight of a baby that has a twin. The mean birth weight of a random sample of 85 single-birth babies is 3086 grams. Assume the population standard deviation is 563 grams. The mean birth weight of a random sample of 68 babies that have a twin is 2263 grams. Assume the population standard deviation is 624 grams. At α=0.10, can you support the pediatrician’s claim? Interpret the decision in the context of the original claim.

Accepted Answer

Hello everyone. Let's take a look at this question together. A school psychologist believes that the average test score of students who receive tutoring is higher than those who do not. In a random sample of 55 students, the mean score is 79.5, with a population standard deviation of 8.2. In a sample of 60 students without tutoring, the mean score is 75.1, with a population standard deviation of 9.6. At alpha equals 0.05, does the data support the psychologist's belief? Is it answer choice A, sample size is too small to conclude? Answer choice B, there is insufficient evidence to support the psychologist's belief? Answer choice C, there is sufficient evidence to support the psychologist's belief, or answer choice D, more data is needed. So in order to solve this question, we have to recall how to test the claim in order to determine whether the data supports the psychologist's belief at the alpha equals 0.0 significance level that the average test score of students who receive tutoring is higher than those who do not. And we know we have a random sample of 55 tutored students with a mean score of 79.5 and a population standard deviation of 8.2, as well as a sample of 60 students without tutoring with a means. of 75.1 and a population standard deviation of 9.6. And so the first step in testing this claim to determine if the data supports the psychologist's belief is to state the hypothesis. which we have our null hypothesis, which is population mean 1, is less than or equal to population mean 2. And our alternative hypothesis, which is population mean 1, is greater than population mean 2. And in our hypotheses, we know that mean 1 is the mean score for tutored students, and mean 2 is the mean score for non-tutored students. And so we calculate our test statistic using the test statistic formula. Which is Z is equal to in parentheses sample mean 1 subtracted by sample mean 2, which is then subtracted by 0 divided by the square root of our population variance 1 divided by sample size 1 plus population variance 2 divided by sample size 2, and then we substitute the values into the formula for the test statistic, which gives us our Z is approximately 2.65. And then we determine the critical value for alpha equals 0.05, which this is a right tail test. So our critical Z value is approximately 1.645. And then we compare our test statistic to the critical value. And since our test statistic of 2.65 is greater than our critical value of 1.645, we reject the null hypothesis. Therefore, we can conclude that there is sufficient evidence at alpha equals 0.05 to support the psychologist's belief. And looking at our answer choices, we can identify that answer choice C is the correct answer, as there is sufficient evidence to support the psychologist's belief. That the average test score of students who receive tutoring is higher than those who do not. So answer choice C is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

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