For many years the standard for the mean weight of a newborn African Elephant ha
ID: 3318165 • Letter: F
Question
For many years the standard for the mean weight of a newborn African Elephant has been thought to be 234 pounds. Elephant weights are normally distributed. Has the average weight decreased due to environmental stress? That is the question. Angel Sonny, the elephant ranger in Kruger National Park, randomly selected 15 newborn African elephants and weighed them. The weights in pounds were as follows:
187, 251, 242, 241, 183, 187, 194, 280, 238, 191, 251, 229, 275, 178, 249
Let the true (unknown) mean newborn elephant weight be with a true (unknown) standard deviation of . We want to see if < 234 pounds.
The null hypothesis is H0:=234. We will test this against the alternative Ha.
If we conclude that is < 234 then Angel will get a huge opportunity to reverse things. This opportunity should be well-deserved (not just lucky) so we do not want to reject H0 unless we are pretty sure that < 234.
Let xbar = the sample mean and s = the sample standard deviation.
We want to test at the 4% level.
f) Calculate the critical value, tstar, for your test.
h) Calculate the p-value for this test.
j) If we ran 10000 4% level tests then about how many times would we make a Type I error?
k) Create a 96% confidence interval for using this data.
Explanation / Answer
Given that,
population mean(u)=210
sample mean, x =240.2
standard deviation, s =54.27
number (n)=15
null, Ho: =210
alternate, H1: >210
level of significance, = 0.04
from standard normal table,right tailed t /2 =1.887
since our test is right-tailed
reject Ho, if to > 1.887
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =240.2-210/(54.27/sqrt(15))
to =2.1552
| to | =2.1552
critical value
the value of |t | with n-1 = 14 d.f is 1.887
we got |to| =2.1552 & | t | =1.887
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.1552 ) = 0.02452
hence value of p0.04 > 0.02452,here we reject Ho
ANSWERS
---------------
f.
null, Ho: =210
alternate, H1: >210
test statistic: 2.1552
critical value: 1.887
decision: reject Ho
h.
p-value: 0.02452
k.
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 = [ 225.0667 ± t a/2 ( 35.0112/ Sqrt ( 15) ]
= [ 225.0667-(1.887 * 9.04) , 225.0667+(1.887 * 9.04) ]
= [ 208.008 , 242.125 ]
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interpretations:
1) we are 96% sure that the interval [ 208.008 , 242.125 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 96% of these intervals will contains the true population mean
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