Answer:
test statistic (Z) is 2.5767 and p-value of the test is .009975
Step-by-step explanation:
: percentage of students who smoke did not change
: percentage of students who smoke has changed
z-statistic for the sample proportion can be calculated as follows:
z=
where
- p(s) is the sample proportion of smoking students (
=0.25)
- p is the proportion of smoking students in the survey conducted five years ago (18% or 0.18)
- N is the sample size (200)
Then, z=
≈ 2.5767
What is being surveyed is if the percentage of students who smoke has changed over the last five years, therefore we need to seek two tailed p-value, which is .009975.
This p value is significant at 99% confidence level. Since .009975 <α/2=0.005, there is significant evidence that the percentage of students who smoke has changed over the last five years
We will just do the following steps:-
389 - 40 = 349
349 - 30 = 319
So, in the end your class had 319 milk jugs left.
Hope I helped ya!!
Answer:
(120 * 32) + 160,500
Step-by-step explanation:
Brainliest!
I don’t know if I’m wrong but Yeah I Think It’s C
Answer: B) 4.963±0.019.
Step-by-step explanation:
Confidence interval for population mean ( when population standard deviation is not given) is given by :-
, where 
= Sample mean
n= Sample size
s= sample standard deviation
t* = critical t-value.
As per given:
n= 50
Degree of freedom = n-1 =49
s= 0.067 lb
For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.
Now , substitute all values in the formula , we get
Hence, a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is 
.
Thus , the correct answer is B) 4.963±0.019.
