Mathematics: The average weight of a package of rolled oats is supposed to be at least 18 ounces. a sample of 18 packages shows a mean of 17.78 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. (b) is this conclusion sensitive to the choice of a? (c) use excel to find the p-value. interpret it.

Options:(A)x = StartFraction 6 over 4 EndFraction (B)x = Negative StartFraction 6 over 4 EndFraction (C)x minus one-fourth = negative StartFraction 5 over 4 EndFraction (D)x = negative three-halves (E)x = negative three-fourths(B)x = Negative StartFraction 6 over 4 EndFraction (D)x = negative three-halves Step-by-step explanation:We want to determine which fraction is equivalent to This value of x is the result in Option B.Reducing to its lowest form: which is Option D.Therefore, the correct options...

The average weight of a package of rolled oats

Ответ:

atiyawhite7863

11.01.2022

Mathematics

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, $\bar{x}$ = 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

$H_{0}: \mu = 18\text{ ounces}\\H_A: \mu < 18\text{ ounces}$

We use one-tailed(left) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{17.78 -18}{\frac{0.41}{\sqrt{18}} } = -2.276$

Now, $t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom for one tail } = -1.739$

Since,

$t_{stat} < t_{critical}$

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.