Question

●A traditional manufacturing process has produced millions of TV tubes with a mean life 1200h and st. deviation 300h. The engineering department of the company introduced a new process. A sample of 100 tubes from new process gives sample mean 1265h. Assuming the st. deviation of new process is same as traditional process, test the following 66 deviation of new process is same as traditional process, test the following hypothesisat5%significantlevel.

1. Traditional method and new method gives same mean life

2. New method is better than the traditional method

3. Traditional method is better than the new method

Answer 1

SOLUTION1 :NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

level of significance=0.05

Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is zc​=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=2.167>zc​=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0303, and since p=0.0303<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 1200, at the 0.05 significance level.

SOLUTION 2:(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=1200

Ha: μ<1200

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is 5α=0.05, and the critical value for a left-tailed test is zc​=−1.64.

The rejection region for this left-tailed test is R={z:z<−1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=2.167≥zc​=−1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9849, and since p=0.9849≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is less than 1200, at the 0.05 significance level.

SOLUTION 3) (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=1200

Ha: μ>1200

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis: Since it is observed that z=2.167>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0151, and since p=0.0151<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 1200, at the 0.05 significance level.