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)
The answers are:
1. 35 burgers
2. 196 cheeseburgers
3. 630 burgers
4. $3.20
Answer:
Hi! The answer to your question is 0 8/25
Step-by-step explanation:
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Hope this helps!!
- Brooklynn Deka
Answer:
a. quotient refers to division
I am not in 8th grade but it might be the 2nd one
