D. one kelvin is the answer
I think the group 2 survey is likely to be biased towards people who are not too keen on zoos since all the values are small.
From group 1 and 3 the mean number of visits are
(4 + 10 + 6 + 3 + 8 + 7 + 9 + 3 + 4 + 7) / 10
= 61/10
mean = 6
Answer:
P = 0.0438 ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.
Due to the small sample size.
Step-by-step explanation:
H0: μ = 500
H1 : μ > 500
Test statistic :
(xbar - μ) / S.E
Tstatistic = (1000 - 500) / 200
= 500 / 200
= 2.5
Pvalue from Test score ; df = 4 - 1 = 3 ; using calculator :
Pvalue = 0.0438
At α = 0.05
Pvalue < α ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.
The test above has a very small sample size, and for a distribution to be approximately Normal, the sample size must be sufficiently large enough according to the Central limit theorem.
For a two sided analysis ; the Pvalue is twice that for the one sided, hence, Pvalue = (0.0438 * 2) = 0.0876 yielding a less strong evidence against the Null.
Answer:
Step-by-step explanation:
What's the question?
Answer:
3x+2x=20
5x=20
x=4
Tyler's number is 4.
Step-by-step explanation:
