Respuesta :
Answer:
We conclude that the mean is smaller than the specifications.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 18 ounces
Sample mean, [tex]\bar{x}[/tex] = 17.78 ounces
Sample size, n = 18
Alpha, α = 0.05
Sample standard deviation, s = 0.41 ounces
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 18\text{ ounces}\\H_A: \mu < 18\text{ ounces}[/tex]
We use one-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{17.78 -18}{\frac{0.41}{\sqrt{18}} } = -2.276[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom for one tail } = -1.739[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to accept the null hypothesis and reject it and conclude that the mean is smaller than the specifications.
b) Yes, the conclusion is sensitive to the choice of null hypothesis.
c) P-value: 0.018036
Since the p value is lower than the significance level, we reject the null hypothesis.
The p-value tells us the probability of null hypothesis being true. It gives the probability of the observed value being true when the null hypothesis is true. Thus, 0.018036 is the probability that the 18 packages shows a mean of 17.78 ounces when the true mean or population mean is 18 ounces.
