A coin-operated coffee machine made by BIG Corporation was designed to discharge

Question

A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of 7.2 ounces of coffee per cup. If it dispenses more than that on average, the corporation may lose money, and if it dispenses less, the customers may complain. Believing that the mean amount of coffee ? dispensed by the machine is greater than 7.2 ounces, BIG plans to do a statistical test of the claim that the machine is working as designed. BIG gathers a random sample of 125 amounts of coffee dispensed by the machine Suppose that the population of amounts of coffee dispensed by the machine has a standard deviation of 0.8 ounces and that BIG performs its hypothesis test using the 0.01 level of significance Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated H : ? IS | less than or equal to 7.2 v What are the null and alternative hypotheses that BIG should use for the test? H : ? is | greater than V7.2 Assuming that the actual value of ? is 7.43 ounces, what is the probability that BIG rejects the null hypothesis? Round your response to at least two decimal places. What is the probability that BIG commits a Type I error? Round your response to at least two decimal places. The probability of committing a Type II error in the second test is greater Suppose that BIG decides to perform another statisticalO test using the same population, alternative hypotheses, and the same sample size, but this second test BIG uses a significance level of 0.1 instead of a significance level of 0.01. Assuming that the O actual value of ? is 7.43 ounces, how does the probability that BIG commits a Type II error in this second test compare to the probability that BIG commits a Type II error in the original test? the same null and committing a Type II error in the second test is less o The probabilities of committing a Type II error are equa

Explanation / Answer

(b) Here standard error of the sample se0 = s/sqrt(n) = ?/sqrt(n) = 0.8/sqrt(125) = 0.07155

test statistic

Z = (x? -?H)/se0 = (7.43 - 7.20)/0.07155 = 3.21

Probability of type I error = Pr(Z > 3.21) = 1 - 0.9993 = 0.0007

Here as we reduce the type I error from 0.01 to 0.1 that will decrease the area of acceptance of null hypothesis and that would increase the probability of rejeting the null hypothesis so that will automatically decrease the probability of type II error and that will increase the power. so Here the probability of committing the type II error in the second test is lesser.

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