Question

The Lubbock County Bureau of Weights and Measures received complaints that a local bottling company was putting less than 20 ounces of root beer in its bottles (labeled 20 oz.). A random sample of fourteen 20-oz. bottles will be randomly selected and its contents will be measured to test the claim that the bottling company is putting significantly less than 20 ounces of root beer in its bottles. Use a 5% level of significance.

9). [2 points] A rival root beer company has 25 of its 20-ounce bottles that are randomly sampled from various stores. The average amount of root beer in each bottle is 19.72 ounces with a standard deviation of 1.28 ounces. Test the claim that the averages from the two companies are equal. (all skills)

for the fist company standard deviation is 0.276 and sample mean is 19.774

Answer 1

Given that,
population mean(u)=20
sample mean, x =19.72
standard deviation, s =0.276
number (n)=25
null, Ho: μ=20
alternate, H1: μ<20
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.711
since our test is left-tailed
reject Ho, if to < -1.711
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =19.72-20/(0.276/sqrt(25))
to =-5.0725
| to | =5.0725
critical value
the value of |t α| with n-1 = 24 d.f is 1.711
we got |to| =5.0725 & | t α | =1.711
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -5.0725 ) = 0.00002
hence value of p0.05 > 0.00002,here we reject Ho
ANSWERS
---------------
null, Ho: μ=20
alternate, H1: μ<20
test statistic: -5.0725
critical value: -1.711
decision: reject Ho
p-value: 0.00002
we have enough evidence to support the claim that the bottling company is putting significantly less than 20 ounces of root beer in its bottles.
9.
Given that,
mean(x)=20
standard deviation , s.d1=0.276
number(n1)=14
y(mean)=19.72
standard deviation, s.d2 =1.28
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =20-19.72/sqrt((0.07618/14)+(1.6384/25))
to =1.051
| to | =1.051
critical value
the value of |t α| with min (n1-1, n2-1) i.e 13 d.f is 2.16
we got |to| = 1.05099 & | t α | = 2.16
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.051 ) = 0.312
hence value of p0.05 < 0.312,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.051
critical value: -2.16 , 2.16
decision: do not reject Ho
p-value: 0.312
we do not have enough evidence to support the claim that the averages from the two companies are equal(skills).