Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
I hope this helps you
2. (1,2,3,4,5,...)
2n
You first find the average of the first two test scores. When you find the average, you add the two numbers together and then divide by two which gives you 89. You then plug in different scores above the number 89 and find the average of those (89 and x) to see which gives you an average of 92. It’s mostly a guess and check type of problem.
Answer: I don't know
Step-by-step explanation: I don't know but have a nice day/night
Answer:
Step-by-step explanation:
I bought 12 pounds of coffee. I used 1/4 of the coffee in a week. How many pounds did i used in a week??
