How to Reject Null Hypothesis

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.