Please answer the following, thanks!
Solution
Given ,
n = 18 .... sample size
= 0.01
This sample is as
0.265,0.262,0.347,0.222,0.181,0.224,0.318,0.268,0.309,0.274,0.196,0.276,0.268,0.313,0.282,0.304,0.226,0.246
The hypothesis are
HO: μ = 0.338 (null hypothesis)
H1 : μ 0.338 (alternative
hypothesis)
Since the population SD is unknown , we use t distribution.
First we need to find the sample mean
and sample SD s.
=
= 0.2656
Now ,
s=
Using given data, find Xi-
for each term.take squre for each.then we can easily find s.
s= 0.04408166
Now the test statistic t is given by
t =
= (0.2656 - 0.338)/(0.04408166/18)
= -6.97
Now we find p value.
n = 18
d.f = n - 1 = 17
tcal = -6.97
Two tailed test
p
value = 0.000
p value is less than
.
So, we do reject the null hypothesis.
Option B . There is sufficient evidence at 0.01 level to conclude that the true mean bubble rising velocity is not 0.338 m/s
Please answer the following, thanks! A study was performed on a method of purifying nuclear fuel...