Answer:
a) No
b) No
c) P value is more than 0.05.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 12 ounces
Sample size, n = 49
P-value = 0.136
First, we design the null and the alternate hypothesis
We use One-tailed z test to perform this hypothesis.
a) Alpha, α = 0.05
Since, p-value > α,
The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.
b) Alpha, α = 0.10
Since, p-value > α,
The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.
c) The evidence is statistically significant at the .05 level means that the p value is more than 0.05.
Dude, what kind of question is this??? Here's my best shot.
If they start at the 32-mile mark, then the we can use 32 as the intercept. 50 can be the change in position over time, so
y=50x+32 (?)
Answer: 28/r+4
Step-by-step explanation: Divide each term by r + 4 and simplify
The population 150 students
The sample is 54% of students
I’m not sure but I hope this helps!
Answer:
I believe it is 54.40$
Step-by-step explanation:
I think weird. so
68 divided by 5 which will give you 20%(13.6) then multiply by 4 which is 80%(54.4)