Question

In a factory where light bulbs are produced, the mean lifespan of a certain type of bulb was 100062
hours with variance 10 hours . However a new and faster production technique was introduced recently and there is some fear that this has reduced the mean lifespan. To investigate this concern, the plan is to measure the lifespan of 200 light bulbs produced with the new technique. If the mean lifespan of these bulbs is 980 hours or less it will be decided that the overall mean lifespan  of new production technique is smaller than 1000 and the new technique will be revised; but if the mean lifespan of the 200 bulbs is more than 980, nothing will be done.
a) Determine the null hypothesis H0 and the alternative hypothesis H1.
b)Determine the statistical procedure (test statistic and rejection region) that apparently is used.
c) i) Calculate the probability that this statistical procedure will not reject H0 while this is incorrect since the true  mean is 990 hours.

Answer 1

The provided sample mean is $\bar{X}=1000$ and the known population standard deviation is $\sigma=10,$ and the sample size is $n=200.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu \geq980$

$H_1:\mu<980$

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

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test is $z_c=-1.64.$

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

The z-statistic is computed as follows:

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}={1000-980 \over 10/\sqrt{200}}=28.284$$

Since it is observed that $z=28.284>-1.64=z_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 980, at the 0.05 significance level. There is not enough evidence to claim that the new technique will be revised.

Using the P-value approach: The p-value is $p=1,$ and since $p=1>0.05,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population $\mu$ is less than 980, at the 0.05 significance level. There is not enough evidence to claim that the new technique will be revised.

c) $\alpha=0.05, z_c=-1.64$

For what values of $\bar{X}$ will we reject $H_0$

$$z_c={\bar{X_c}-\mu \over \sigma/\sqrt{n}}=>\bar{X_c}=\mu+z_c\cdot{\sigma \over \sqrt{n}}=$$

$$=980+(-1.64){10 \over \sqrt{200}}\approx978.84$$

If $\mu=990,$ what is $P(Type\ II\ error)$?

$$P(Type\ II\ error)=P(Do\ not\ reject\ H_0|\mu=990)=$$

$$=P(\bar{X}>978.84|\mu=990)=P(Z>{978.84-990 \over 10/\sqrt{200}})=$$

$$=P(Z>-15.7826)=1=\beta$$

$$Power\ of\ the\ test=1-\beta =1-0=0$$

The probability that this statistical procedure will not reject $H_0$ while this is incorrect since the true mean $\mu$ is 990 hours equals to $0.$