Complies. 2. To determine whether the pipe welds in a nuclear power plant meet
... Weld strength is measured as the force required to break ... calculate a single
tail probability for this one-sided test, using the probability distribution with mean.
Hypothesis test problems 1. For the following pairs of assertions, indicate which do not comply with rules for setting up hypotheses, and why. 1.1. H 0 : 100 , H a : 100 Complies 1.2. H 0 : 20 , H a : 100
Doesn’t comply because alternative doesn’t contradict null.
Alternative has an equal to, and so does the null. 1.3. H 0 : 1 2 25 , H a : 1 2 100 The alternative must be the exact opposite of the null, or, if it is one sided, it should be the exact opposite of the null if the null had the opposite inequality. In other words, if the 100 were a 25, this would be OK. 1.4. H 0 : S12 S 22 , H a : S12 S 22 Does not comply because these are statements about sample quantities, not parameters. 1.5. H 0 :
1
1 2 1 , H a : 2 1
Complies.
2. To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests (not hypothesis test, but tests of strength of the welds) are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose specifications state that the mean strength of welds should exceed 100 lb/in2. What are the appropriate null and alternative hypotheses? We want strong evidence before we accept that the mean weld strength is greater than 100 lb/in2. Thus, letting represent the true mean weld strength for the conceptual population of pipes in a nuclear power plant, the alternative hypothesis will be H a : 100
H : 100
, and the null will be the
H : 100
, which we re‐write as an equal to— 0 —so that we can opposite of this, 0 calculate a single tail probability for this one‐sided test, using the probability distribution with mean 100 . 3. Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40‐amp fuses wants to make sure that the mean amperage at which fuses burn out is actually 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40, the manufacturer may be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample is to be selected and inspected. What null and alternative hypotheses are of interest to the manufacturer? Here, the manufacturer will not change the way the nominal (that means the fuses that are names this way or labeled this way) 40‐amp fuses are made unless there is strong evidence that they do not have the nominal mean amperage, and there are negative consequences if the mean is below or
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above the nominal mean. Let represent the true mean amperage of these fuses. The alternative hypothesis is then H a : 40
, and the null is the opposite of this:
H 0 : 40 .
4. Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most 150 degrees F, there will be no negative effects on the river’s ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge‐water temperature above 150 degrees, 50 water samples will be taken at randomly selected times, and the temperature of each sample recorded. What are the appropriate null and alternative hypotheses? Here, we want strong evidence before we accept that the water is safe, because assuming the water is safe when it isn’t safe would be a mistake with pretty bad consequences. Thus, if represents the true mean temperature of the discharged water, we want H1 : 150
H : 150
, and the null will be
H : 150
, but then we change it to be an equal to, 0 , so that the opposite of this, 0 we can calculate probabilities with mean 150. 5. A mixture of pulverized ash and Portland cement to be used for grouting should have a compressive strength of more than 1300 KN/m3. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met at a significance level of .05 . Suppose compressive strength for specimens of this mixture is normally distributed with 60 . Let denote the true average compressive strength. 5.1. What are the appropriate null and alternative hypotheses?
H 0 : 1300 and H a : 1300 5.2. Suppose the compressive strength of n 20 randomly selected specimens is measured, and the sample mean is x 1320 . Calculate the p‐value. .068 5.3. At the chosen significance level, should the mixture be used? State your answer with a complete sentence. If the true compressive strength of this mixture is 1300 KN/m3, the probability that we would observe a sample mean of 1320 KN/m3, when the sample size is 20, is .068. Since this probability is not lower than the pre‐chosen significance level of .05 , we cannot reject the hypothesis that the strength is 1300, in favor of the alternative that it is strong enough to meet specifications. In short, no, the mixture should not be used. 5.4. This test can be performed by calculating which type of 95% confidence interval? 5.4.1.A symmetric interval. 5.4.2.A lower bound. This one. 5.4.3.An upper bound. 5.5. Calculate the interval and show that you get the same result.
1297 . x 1.645 60 20 Page 2 of 4
We’re 95% confident that the true compressive strength of the mixture is above 1297 KN/m3. Since this interval contains 1300, we cannot reject the null. 6. The calibration of a scale is to be checked by weighing a 10‐kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed. Let denote the true average weight reading on the scale. 6.1. What hypotheses should be tested? H 0 : 10 H a : 10 , . 6.2. Assume a significance level of .01 . Calculate the p‐value and state your conclusion based on the following data. 9.981 10.006 9.857 10.107 9.888 9.728 10.439 10.214 10.190 9.793 p‐value from TI83, using data as input, not statistics, but giving it the sigma from above: .7482 The probability that a sample of 10 weighings would give the results in the above set of data is .7482 if the null is true. The data above are very much consistent with the null, at the significance level .01 (and at lots of higher ones) so that we cannot reject the null. 6.3. This test can be performed by calculating which type of 99% confidence interval? 6.3.1.A symmetric interval.—this one. 6.3.2.A lower bound. 6.3.3.An upper bound. 6.4. Calculate the interval and show that you get the same result. You can use your calculator to do a 2‐sided interval: STAT, TEST, Z‐interval, tell it you’re using data, give it sigma, tell it which list has the data, freq = 1, give it C, calculate… We are 99% confident that the mean mass of the weight as measured by this scale falls between 9.857 and 10.183 kg. Since this interval contains 10, we cannot reject the null.
6.5. In the context of this problem, write a sentence that gives the most common misinterpretation of a p‐value. The probability that the null hypothesis is true is .7482. Or The probability that 10 .7482 is . Or .7482 The probability that the true mean mass as measured by this scale is 10 kg is . Page 3 of 4
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