MUST use MiniTab A chemist claims that the average length of a Python in Kingsto

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MUST use MiniTab

A chemist claims that the average length of a Python in Kingston`s River is larger than
25.2m. The lengths of 20 randomly selected salmons from the river are below:-

26.03 25.73 25.23 25.44 25.69 26.42 24.84 25.2 25.68 25.63
24.75 24.55 24.87 25.56 25.36 25.5 25.49 24.99 25.39 25.87

1) Calculate the sample mean and standard deviation
2) Assume that the length of a salmon fish is normally distributed and test the claim at
a significance level of 0.05 without using the p-value concept. Clearly show the 8
steps of your hypothesis test.
3) Calculate the confidence interval limits using the p-value concept

Explanation / Answer

Given that,
population mean(u)=25.2
sample mean, x =25.411
standard deviation, s =0.459
number (n)=20
null, Ho: <25.2
alternate, H1: >25.2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.729
since our test is right-tailed
reject Ho, if to > 1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25.411-25.2/(0.459/sqrt(20))
to =2.0558
| to | =2.0558
critical value
the value of |t | with n-1 = 19 d.f is 1.729
we got |to| =2.0558 & | t | =1.729
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.0558 ) = 0.0269
hence value of p0.05 > 0.0269,here we reject Ho
ANSWERS
---------------
null, Ho: =25.2
alternate, H1: >25.2
test statistic: 2.0558
critical value: 1.729
decision: reject Ho
p-value: 0.0269

have evidence that it is greater than 25.2

given that,
sample mean, x =25.411
standard deviation, s =0.459
sample size, n =20
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 19 d.f is 1.729
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 25.411 ± t a/2 ( 0.459/ Sqrt ( 20) ]
= [ 25.411-(1.729 * 0.103) , 25.411+(1.729 * 0.103) ]
= [ 25.234 , 25.588 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 25.234 , 25.588 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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