A machine that is programmed to package 2.55 pounds of cereal in each cereal box
ID: 3242477 • Letter: A
Question
A machine that is programmed to package 2.55 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 25 cereal boxes, the mean and standard deviation are calculated as 2.58 pounds and 0.10 pound, respectively. Use Table 2.
Select the null and the alternative hypotheses to determine if the machine is working improperly, that is, it is either underfilling or overfilling the cereal boxes.
Calculate the value of the test statistic. (Round your answer to 2 decimal places.)
A machine that is programmed to package 2.55 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 25 cereal boxes, the mean and standard deviation are calculated as 2.58 pounds and 0.10 pound, respectively. Use Table 2.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
a) Null hypothesis: = 2.55
Alternative hypothesis: 2.55
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 = 0.02
DF = n - 1 = 25 - 1
D.F = 24
b) t = (x - ) / SE
t = 1.50
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 -1.77 or greater than 1.77.
c -1) Thus, the P-value = 0.1336
0.10< p-value < 0.20
c-2) Do not reject H0 since the p-value is greater than .
Interpret results. Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.
d -1) The critical value(s) at a 5% level of significance.
d- 2) Yes, the machine is working improperly.
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