QUESTION 1
Because
of high production-changeover time and costs, a director of manufacturing must
convince management that a proposed manufacturing method reduces cost before
the new method can be implemented. The current production method operates with
a mean cost of $220 per hour. A research study will measure the cost of the new
method over a sample production period.
a. Develop
the null and alternative hypothesis most appropriate for this study.
b. Comment
on the conclusion when H0 cannot be rejected.
H0
cannot be rejected if the confidence interval include the hypothesized value.
It mean that if p-value larger than the significance level, H0
is not rejected. When H0 cannot be rejected,
new method can be implemented.
c. Comment
on the conclusion when H0 can be rejected.
H0
is rejected if the confidence interval does not include the hypothesized value.
It mean that if p-value smaller than the significance level, H0
is rejected. When H0 can be rejected new
method cannot be implemented.
|
Does
not reject H0
|
Rejects
H0
|
H0
True
|
√
|
Type
׀ error
|
H0
False
|
Type
׀׀ error
|
√
|
d. What
is the Type ׀
Error in this situation? What are the consequences of making this error?
Type ׀ error in this situation is reject the null
hypothesis, H0 when it is true, consequences the method
cannot be implemented.
e.
What is the Type ׀׀ Error in this situation? What are the, consequences
of making this error?
Type
׀׀
error in this situation is accept the null hypothesis when it is false, consequences
the method can be implemented.
QUESTION 2
H0:=
15
Ha:≠
15
a.
Compute the value of
the test
b.
What is the what p- Value?
2(0.5-0.47741)-two tailed test (0.4774
from 2.003 in the tables of appendix B.1)
=
0.04518
c.
At α = 0.05, what is your conclusion?
Reject H0 when z does not fall in the region
between -1.96 and 1.96.
H0
is rejected if z < -1.96 or z > 1.96.
In my conclusion, H0 is rejected because
-2.003 fall on the region of rejection, so H0
is rejected. Moreover 0.04518<1.96, it mean that we reject H0
if the p-value is smaller than the significance level. Besides that -2.003 lies
in the area to the left of -1.96 of rejection area, so I conclude that H0
is rejected.
d.
What is the rule using the critical
value? What is your conclusion?
Rule of using the critical value is we reject H0
when
p-value lies in the area to the left or right of the critical value, it mean
that if they lies to the left or right of critical value we reject H0,
because they are in the area of region of rejection. Other rules is, in a one
tailed test we put all the rejection region in one tail, while in two tailed
test, we put the rejection region in two tail, left and right of bell shaped.
In my conclusion H0 is rejected because the
confidence interval does not include the hypothesized value because the value
of z = -2.003 does not include in the interval of -1.96 and 1.96. It is because
-2.003 are smaller than -1.96, and -2.003 fall on the region of rejection, so H0
is rejected, and we conclude that the population mean is different.
QUESTION 3
b. Use
the t distribution table to compute a range for the p-value
Assuming
a 5% significance level, reject H0
if t > 1.771(1.771 from table in
appendix B.2, [n-1,25-1=24]). The computed t value of 2.315 is between 1.318 and
1.711. Therefore, the p-value is between 0.10 and 0.05
c. At
α = 0.05, what is your conclusion?
H0
is rejected because 2.315 >1.771, it is because p-value is smaller than the
significance level, 1.771 is smaller than 2.315, so H0
is rejected.
d. What
is the rejection rule using the critical value? What is your conclusion?
Rule
of using the critical value is we reject H0 when
p-value lies in the area to the left or right of the critical value, it mean
that if they lies to the left or right of critical value we reject H0,
because they are in the area of region of rejection. Other rules is, in a one
tailed test we put all the rejection region in one tail, while in two tailed
test, we put the rejection region in two tail, left and right of bell shaped.
In my conclusion H0 is rejected because the
confidence interval does not include the hypothesized value because the value
of t = 2.315 does not include in the t <1.771, it is because 2.315>1.771
and 2.315 fall on the region of rejection, moreover the computed t of 1.318
lies in the area to the left of 1.711 so H0
is rejected. We accept the Ha and conclude
that the population mean is more than 12 and we conclude that the population
mean is different.
QUESTION 4
n=120
σ = 5
α
= 0.05(one tailed test)
a. If
the population mean is 9, what is the probability that the sample mean leads to
the conclusion do not reject H0?
Reject H0
when z < -1.645 (0.5-0.05=0.45,0.45
from the tables in appendix B.1 became 1.645)
p-value=
0.5– 0.4858(0.4858 from 2.191 in the table appendix B.1)
= 0.0142
The
probability that the sample mean leads to the conclusion do not reject H0
is
0.0142,it is only 1.42%.
b. What type of error
would be made if the actual population mean is 9 and we conclude that H0μ
>_ 10 is true?
The
type of error would be made if the actual population mean is 9 and we conclude
that H0μ
>_ 10 is true is Type ׀
error, it is because we reject the null hypothesis,H0
when
it is true.
c. What
is the probability of making a Type ׀׀ error if the actual population mean is 8.
Type
׀׀
error is accepting the null hypothesis when it is false.The probabilility of
making Type ׀׀
error is zero(0). It mean that we have extremely evidence that H0
is not true.We have extremely evidence to reject H0.
QUESTION 5
The
chamber of commerce of Florida Gulf Coast community advertises that the area
residential property is available at a mean cost of $125000 or less per lot.
Suppose a sample of 32 properties provided a sample mean of $130000 per lot and
a sample standard deviation of $12500. Use a 0.05 level of significance to test
the validity of the advertising claim.
The
test is one tailed because we want to determine whether there has been a
increasing in the cost. The inequality in the alternate hypothesis point to the
region of rejection in the right tail of the distribution. We do not know the
standard deviation of the population. So we substitute the sample standard
deviation. The value of the test statistics is computed by formula (10-2):
Critical value
We reject H0
because 2.2627>1.696, It mean that we reject H0
if the t value are larger than critical value. Moreover 2.2627 lies in the
region to the right of the critical value of 1.696, so the null hypothesis is rejected
at the 0.05 significance level. The computed t of 2.2627 is between 1.309 and
1.696. Therefore, the p-value is between 0.1 and 0.05 We have demonstrated that the cost of advertising measured
increase the mean cost per claim more than $12500. So we conclude that null
hypothesis is rejected because fall on the region of rejection.
