The average weight of a package of rolled oats is supposed to be at least 15 oun
ID: 3437860 • Letter: T
Question
The average weight of a package of rolled oats is supposed to be at least 15 ounces. A sample of 18 packages shows a mean of 14.71 ounces with a standard deviation of 0.41 ounce.
At the 5 percent level of significance, is the true mean smaller than the specification? Clearly state your hypotheses and decision rule.
H0: 15. Reject H0 if tcalc < -1.740
rejected the null hypothesis.
failed to reject the null hypothesis.
(a)At the 5 percent level of significance, is the true mean smaller than the specification? Clearly state your hypotheses and decision rule.
H1: < 15. Reject H1 if tcalc < -1.740 H1: < 15. Reject H1 if tcalc > -1.740 H0: 18. Reject H0 if tcalc > -1.740H0: 15. Reject H0 if tcalc < -1.740
(b) If = 0.01, we would haverejected the null hypothesis.
failed to reject the null hypothesis.
(c) Use Excel to find the p-value. (Round your answer to 4 decimal places.) p-valueExplanation / Answer
The average weight of a package of rolled oats is supposed to be at least 15 ounces. A sample of 18 packages shows a mean of 14.71 ounces with a standard deviation of 0.41 ounce.
(a)
At the 5 percent level of significance, is the true mean smaller than the specification? Clearly state your hypotheses and decision rule.
H1: < 15. Reject H1 if tcalc < -1.740
H1: < 15. Reject H1 if tcalc > -1.740
H0: 18. Reject H0 if tcalc > -1.740
H0: 15. Reject H0 if tcalc < -1.740
(b)
If = 0.01, we would have
rejected the null hypothesis. ( calculated P value 0.004 < 0.01 level)
failed to reject the null hypothesis.
(c)
Use Excel to find the p-value. (Round your answer to 4 decimal places.)
p-value =0.0040
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
15
Level of Significance
0.05
Sample Size
18
Sample Mean
14.71
Sample Standard Deviation
0.41
Intermediate Calculations
Standard Error of the Mean
0.0966
Degrees of Freedom
17
t Test Statistic
-3.0009
Lower-Tail Test
Lower Critical Value
-1.7396
p-Value
0.0040
Reject the null hypothesis
(a)
At the 5 percent level of significance, is the true mean smaller than the specification? Clearly state your hypotheses and decision rule.
H1: < 15. Reject H1 if tcalc < -1.740
H1: < 15. Reject H1 if tcalc > -1.740
H0: 18. Reject H0 if tcalc > -1.740
H0: 15. Reject H0 if tcalc < -1.740
(b)
If = 0.01, we would have
rejected the null hypothesis. ( calculated P value 0.004 < 0.01 level)
failed to reject the null hypothesis.
(c)
Use Excel to find the p-value. (Round your answer to 4 decimal places.)
p-value =0.0040
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