Question

7. A certain variety of rosebush is known to produce a large number of rose blooms during the blooming period. Alaina is a botanist that knows this particular rosebush usually produces 75 roses per rosebush during the blooming season. However, her experience tells her that this figure may be wrong and need to be re-evaluated. She selects 32 of these rosebushes and finds that on average they produce 78 roses during the blooming season with a standard deviation of 12 roses. With ?? = 0.05, test Alaina’s claim that the average number of roses produced by a single rosebush is different from 75 roses.

a. Are you going to use the ??-value or traditional method? (2 points)

b. Build the hypotheses and find the critical value(s), if needed. (8 points)

c. Find the test statistic and the ??-value, if needed. (6 points)

d. Determine whether the null hypothesis is rejected or not rejected. (3 points)

e. Summarize your results within the context of the problem. (6 points)

Answer 1

7.

Given that,
population mean(u)=75
sample mean, x =78
standard deviation, s =12
number (n)=32
null, Ho: μ=75
alternate, H1: μ!=75
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.04
since our test is two-tailed
reject Ho, if to < -2.04 OR if to > 2.04
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =78-75/(12/sqrt(32))
to =1.4142
| to | =1.4142
critical value
the value of |t α| with n-1 = 31 d.f is 2.04
we got |to| =1.4142 & | t α | =2.04
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 != 1.4142 ) = 0.1673
hence value of p0.05 < 0.1673,here we do not reject Ho
ANSWERS
---------------
a.
yes,
going to use the ??-value or traditional method
null, Ho: μ=75
alternate, H1: μ!=75
c.
test statistic: 1.4142
b.
critical value: -2.04 , 2.04
d.
decision: do not reject Ho
p-value: 0.1673
e.
we do not have enough evidence to support the claim that the average number of roses produced by a single rosebush is different from 75 roses.