Answer to Question #149085 in Statistics and Probability for yedii

Question

company produces low-energy light bulbs. The bulbs are described as using 9 watts of power. Amir, the production manager, asked Jenny to measure the power used by each of a sample of 120 bulbs from the latest batch produced and to test the hypothesis that the mean for the batch is 9.0 watts. Jenny is to carry out the test at the 5% significance level. (a) What assumption must be made about the sample of 120 bulbs if the result of this test is to be valid? (b) Jenny found that the mean for the sample was 9.2 watts and that the standard deviation was 1.3 watts. Carry out the test asked for by Amir. (c) When Jenny reported the conclusion of the test to Amir, he said that he had intended to ask Jenny to test whether the bulbs in the batch use more than 9.0 watts on average. For the test in part (b), state the effect that this new information would have on: (i) the alternative hypothesis; (ii) the critical value(s); (iii) the conclusion.

Answer 1

a) The central limit theorem states that the sampling distribution of the mean approaches a normal distribution, as the sample size increases. This fact holds especially true for sample sizes over 30.

When the population standard deviation is unknown, the mean has a Student's t distribution.

(b) The provided sample mean is $\bar{X}=9.2$ and the sample standard deviation is $s=1.3,$ and the sample size is $n=120.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=9$

$H_1:\mu\not=9$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The number of degrees of freedom are $df=120-1=119,$ and the significance level is  $\alpha=0.05.$ The critical value for a two-tailed test is $t_c=1.9801.$

The rejection region for this two-tailed test is $R=\{t:|t|>1.9801\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{9.2-9}{1.3/\sqrt{120}}=1.6853$$

Since it is observed that $|t|=1.6853<1.9801=t_c,$ 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 different than 9, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.094551,$ and since $p=0.094551>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 9, at the 0.05 significance level.

Therefore, there is not enough evidence to claim that the mean for the batch is not 9.0 watts , at the 0.05 significance level.

(c) The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq9$

$H_1:\mu>9$

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

The number of degrees of freedom are $df=120-1=119,$ and the significance level is  $\alpha=0.05.$ The critical value for a right-tailed test is $t_c=1.657759.$

The rejection region for this right-tailed test is $R=\{t:t>1.657759\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{9.2-9}{1.3/\sqrt{120}}=1.6853$$

Since it is observed that $t=1.6853>1.657759=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 9, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.047276,$ and since $p=0.047276<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 9, at the 0.05 significance level.

Therefore, there is enough evidence to claim that the the bulbs in the batch use more than 9.0 watts on average, at the 0.05 significance level.