The average weight of a package of rolled oats is supposed to be at least 16 ounces. A sample of 18 packages shows a mean of 15.81 ounces with a standard deviation of .48 ounce. |
(a) |
At the 5 percent level of significance, is the true mean smaller than the specification? Clearly state your hypotheses and decision rule. |
a. | H0: μ ≥ 16. Reject H0 if p > 0.05 |
b. | H1: μ < 16. Reject H1 if p < 0.05 |
c. | H0: μ ≥ 16. Reject H0 if p < 0.05 |
d. | H1: μ < 16. Reject H1 if p > 0.05 |
|
(b) |
If α = .010, we would have |
a. | failed to reject the null hypothesis. |
b. | rejected the null hypothesis. |
|
(c) | Use Excel to find the p-value. (Round your answer to 4 decimal places.) |
p-value |
a)
H0: μ ≥ 16. Reject H0 if p < 0.05
b)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 16
Alternative Hypothesis, Ha: μ < 16
Rejection Region
This is left tailed test, for α = 0.05 and df = 17
Critical value of t is -1.74.
Hence reject H0 if t < -1.74
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (15.81 - 16)/(0.48/sqrt(18))
t = -1.679
P-value Approach
P-value = 0.0557
As P-value >= 0.01, fail to reject null hypothesis.
c)
p value = =T.DIST(-1.679,17,TRUE)
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