Question

Two competing household paint companies claim to have quick-drying paint. An experiment was performed to compare the mean lengths of time required for the paints to dry for brand A and brand B. Twelve randomly selected 10 x 10 squares of drywall were painted with brand A. Another 12 were randomly selected and painted with brand B. The lengths of time in minutes for the paint to dry were recorded. The means, standard deviations, and sizes of the two samples follow.

Brand A: x₁ = 21.8 min; s₁ = 8.7 min; n₁ = 12

Brand B: x₂ = 18.9 min; s₂ = 7.5 min; n₂ = 12

Past experience with the paint composition of the two paints permits us to assume that both distributions are approximately normal. Use a 5% level of significance to test the claim that there is no difference between the two brands in the mean time required for drying. Find the sample test statistic.

Answer 1

Given that,
mean(x)=21.8
standard deviation , s.d1=8.7
number(n1)=12
y(mean)=18.9
standard deviation, s.d2 =7.5
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =21.8-18.9/sqrt((75.69/12)+(56.25/12))
to =0.8746
| to | =0.8746
critical value
the value of |t α| with min (n1-1, n2-1) i.e 11 d.f is 2.201
we got |to| = 0.87458 & | t α | = 2.201
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 != 0.8746 ) = 0.4
hence value of p0.05 < 0.4,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.8746
critical value: -2.201 , 2.201
decision: do not reject Ho
p-value: 0.4
we do not have enough evidence to support the claim that there is no difference between the two brands in the mean time required for drying.