A random sample of 18 turkeys was given a new diet over a period of several mont

Question

A random sample of 18 turkeys was given a new diet over a period of several months, and the weight gain in pounds was recoded at the end. The observed mean was 3.7 with standard deviation .84. (a) With 95% confidence, what would the mean weight gain be if all (similar) turkeys were given this diet for this period of time? (b) If a standard diet normally produces a mean weight gain of 3.5 pounds, is there significant evidence that the new diet produces a greater mean weight gain? Why or why not? (c) If the population standard deviation were .84, how many turkeys would need to be sampled to be able to estimate the population mean weight gain with an error no more than .3 and with 95% confidence?

Explanation / Answer

a.
Confidence Interval
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
Mean(x)=3.7
Standard deviation( sd )=0.84
Sample Size(n)=18
Confidence Interval = [ 3.7 ± t a/2 ( 0.84/ Sqrt ( 18) ) ]
= [ 3.7 - 2.11 * (0.198) , 3.7 + 2.11 * (0.198) ]
= [ 3.282,4.118 ]
b.
Given that,
population mean(u)=3.5
sample mean, x =3.7
standard deviation, s =0.84
number (n)=18
null, Ho: <3.5
alternate, H1: >3.5
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.74
since our test is right-tailed
reject Ho, if to > 1.74
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.7-3.5/(0.84/sqrt(18))
to =1.01
| to | =1.01
critical value
the value of |t | with n-1 = 17 d.f is 1.74
we got |to| =1.01 & | t | =1.74
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.0102 ) = 0.16329
hence value of p0.05 < 0.16329,here we do not reject Ho
ANSWERS
---------------
null, Ho: =3.5
alternate, H1: >3.5
test statistic: 1.01
critical value: 1.74
decision: do not reject Ho
p-value: 0.16329
no evidence that the new diet produces a greater mean weight gain
c.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.84
ME =0.3
n = ( 1.96*0.84/0.3) ^2
= (1.65/0.3 ) ^2
= 30.12 ~ 31      

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