Question

1) null hypothesis

2) alternative hypothesis

3) where the region of rejection lies (upper tail, lower tail, both tails)

4) the test that is to be used

5) the degrees of freedom

6) the critical value of the test statistic

7) the computed value of the test statistic

8) the statistical decision (whether the null hypothesis is rejected or not)

9) the p-value

10) the assumptions you made in your work

Now, finally - Here is the question:

Trail mix is sold in individual pouches labeled as containing 8 ounces by weight. The weights of 40 pouches are measured (using a tared scale so that only the weight of the trail mix is measured). The sample mean is found to be 8.5 ounces. The sample standard deviation is calculated to be 0.4 ounces. At 95% confidence, is there evidence that the mean weight per pouch is different from 8 ounces?

Answer 1

Given that,
population mean(u)=8
sample mean, x =8.5
standard deviation, s =0.4
number (n)=40
null, Ho: μ=8
alternate, H1: μ!=8
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.02
since our test is two-tailed
reject Ho, if to < -2.02 OR if to > 2.02
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.5-8/(0.4/sqrt(40))
to =7.906
| to | =7.906
critical value
the value of |t α| with n-1 = 39 d.f is 2.02
we got |to| =7.906 & | t α | =2.02
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 7.9057 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
1.
null, Ho: μ=8
2.
alternate, H1: μ!=8
3.
two tailed test
4.
t test for single mean
5.
degree of freedom d.f is 2.02
6.
test statistic: 7.906
7.
critical value: -2.02 , 2.02
8.
decision: reject Ho
9.
p-value: 0
10.
we have enough evidence to support the claim that the mean weight per pouch is different from 8 ounces.