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.

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, =  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

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

Formula:

Putting all the values, we have

Now,

Since,                  

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.

o= 4(i) +3

step-by-step explanation:

o is your output value and i is your input value. so, for example, if we stuck 7 in as our input, we would first multiply by 4 ( pemdas ), which equals 28. then we would add 3 to 28, for a total output of 31. this may not be a graph, but, it is a way to solve your formula..

answer: kljdcdhjbdlchjad oac apoich a

step-by-step explanation: