It\'s well established, we\'ll assume, that lab rats require an average of 32 tr
ID: 3320764 • Letter: I
Question
It's well established, we'll assume, that lab rats require an average of 32 trails in a complex water maze before reaching a learning criterion of three consecutive errorless trials. To determine whether a mildly adverse stimulus has any effect on performance, a sample of seven lab rats were given a mild electrical shock just before each trial. (A). Given that x=34.89 and s=3.02, test the null hypothesis with t, using the .05 level of significance. (B). Construct a 95 percent confidence interval for the true number of trials required to learn the water maze. (C). Interpret this confidence interval.
Explanation / Answer
a.
Given that,
population mean(u)=32
sample mean, x =34.89
standard deviation, s =3.02
number (n)=7
null, Ho: =32
alternate, H1: !=32
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =34.89-32/(3.02/sqrt(7))
to =2.532
| to | =2.532
critical value
the value of |t | with n-1 = 6 d.f is 2.447
we got |to| =2.532 & | t | =2.447
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.5319 ) = 0.0446
hence value of p0.05 > 0.0446,here we reject Ho
ANSWERS
---------------
null, Ho: =32
alternate, H1: !=32
test statistic: 2.532
critical value: -2.447 , 2.447
decision: reject Ho
p-value: 0.0446
b.
TRADITIONAL METHOD
given that,
sample mean, x =34.89
standard deviation, s =3.02
sample size, n =7
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.02/ sqrt ( 7) )
= 1.141
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 6 d.f is 2.447
margin of error = 2.447 * 1.141
= 2.793
III.
CI = x ± margin of error
confidence interval = [ 34.89 ± 2.793 ]
= [ 32.097 , 37.683 ]
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DIRECT METHOD
given that,
sample mean, x =34.89
standard deviation, s =3.02
sample size, n =7
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 6 d.f is 2.447
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 = [ 34.89 ± t a/2 ( 3.02/ Sqrt ( 7) ]
= [ 34.89-(2.447 * 1.141) , 34.89+(2.447 * 1.141) ]
= [ 32.097 , 37.683 ]
-----------------------------------------------------------------------------------------------
c.
interpretations:
1) we are 95% sure that the interval [ 32.097 , 37.683 ] 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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