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A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 mg claimed by the manufacturer. Test this claim at the 0.02 level of significance. The mean acetaminophen content for a random sample of n = 41 tablets is603.3 mg. Assume that the population standard deviation is 4.9 mg.
A. Since the test statistic is greater than the critical z, there is sufficient evidence to accept the null hypothesis and to support the claim that the mean content of acetaminophen is 600 mg.
B. Since the test statistic is greater than the critical z, there is sufficient evidence to reject the null hypothesis and to support the claim that the mean content of acetaminophen is not 600 mg.
C. Since the test statistic is less than the critical z, there is sufficient evidence to reject the null hypothesis and to support the claim that the mean content of acetaminophen is not 600 mg.
D. Since the test statistic is greater than the critical z, there is insufficient evidence to reject the null hypothesis and to support the claim that the mean content of acetaminophen is not 600 mg.
A vending machine that dispenses coffee into cups must fill the cups with 7.8oz of liquid. Before selling the vending machine to a college or business, the company tests the machine to be sure it is dispensing an average amount of 7.8oz of coffee.
We took a random sample of 25 customers and found a mean waiting time of 22 minutes with a standard deviation of 6 minutes. We want to construct a 99% confidence interval for the population mean of such waiting times.
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a high school math teacher gave her 30 students a math exam. in this exam the average scores for the high school
However, administrators concerned about cost argued that with only 200,000 trials, the probability of waking up using the new method could not be accurately estimated. Verify that a 0.95 confidence interval for p, the probability of waking up, is ..
It is a striking fact that the first digits of numbers in legitimate records often follow a distribution known as Benford's Law, shown below.
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