The cherry canning company (ccc) of traverse city, Michigan has introduced a new
ID: 3296862 • Letter: T
Question
The cherry canning company (ccc) of traverse city, Michigan has introduced a new manufacturing method to streamline the canning process of cherries. Although the time to fill a can has been reduced the quality control manager is concerned about the uniformity of the amount of cherries in each can. The cans are labled to contain 14.5 ounces of cherries and natural juice to be sure that this level has not been affected by the new method the manager randomly sample 25 cans over an eight hours shift. The mean number of ounces in this sample is 14.66 with a standard deviation of 0.4 ounces. Has the new method changed the amount of cherries that should be in the can? (a) State the null and alternative hypotheses for this experiment. (b) What is the Type I error for this particular application and what are its consequenes? (c) What is the Type II error for this particulars application and what are its consequences?Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 14.5
Alternative hypothesis: 14.5
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 = 0.08
DF = n - 1 = 25 - 1
D.F = 24
t = (x - ) / SE
t = 2.0
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 24 degrees of freedom is less than - 2.0 or greater than 2.0.
Thus, the P-value = 0.057
Interpret results. Since the P-value (0.057) is greater than the significance level (0.05), we cannot reject the null hypothesis.
b) The type I error involves involves the rejection of a null hypothesis that is actually true. In this example the probability of type I error is 0.05.
c) Type II error occurs when we do not reject a null hypothesis that is false.
IN this case if actually the mean is not equal to 14.50, then also we are not able to reject the claim that mean is 14.50. This is type II error.
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