A scale is to be calibrated by weighing a 1200g test weight 50 times. The 50 sca
ID: 3231998 • Letter: A
Question
A scale is to be calibrated by weighing a 1200g test weight 50 times. The 50 scale readings have mean 1200.8g and standard deviation 2g. Find the P-value for testing H_0: mu = 1200 versus H_1: mu 1200. Specifications for a water pipe call for a mean breaking strength mu of more than 2500 lb per linear foot. Engineers will perform a hypothesis test to decide whether or not to use a certain kind of pipe. They will select a random sample of 1 ft sections of pipe, measure their breaking strengths, and perform a hypothesis test The pipe will not be used unless the engineers can conclude that mu > 2500. What would be appropriate null and alternative hypotheses? Industrial firms often employ methods of "risk transfer, " such as insurance or indemnity clauses in contracts, as a technique management. An article reports the results of a survey in which managers were asked which methods played a major role in the risk management strategy of their firms. In a sample of 43 oil companies, 22 indicated that risk transfer played a major role, while in a sample of 90 construction companies, 55 reported that risk transfer played a major role, (These figures were read from a graph.) Can we conclude that the proportion of oil companies that employ the method of risk transfer is less than the proportion of construction companies that do?Explanation / Answer
Solution:-
7)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 1200
Alternative hypothesis: 1200
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.2828
DF = n - 1 = 50 - 1
D.F = 49
t = (x - ) / SE
t = 2.83
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than - 2.83 or greater than 2.83.
Thus, the P-value = 0.00673
Interpret results. Since the P-value (0.00673) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that mean is different from 1200g.
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