Answer to Question #160114 in Statistics and Probability for Gabriel

Question

A factory manufacturing light-emitting diode (LED) bulb claims that their light bulbs last for 50,000 hours on the average. To confirm if this claim was valid, a quality control manager got a sample of 50 LED bulbs and obtained a mean lifespan of 40,000 hours. The standard deviation of the manufacturing process is 1000 hours. Do you think that the claim of the manufacturer is valid? Use the 95% confidence level.

Answer 1

The provided sample mean is $\bar{x}=40000$ and the sample standard deviation is $s=1000.$

The sample size is $n=50.$ The number of degrees of freedom are $df=n-1=49,$ and the significance level is $\alpha=0.05.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=50000$

$H_1: \mu\not =50000$

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

Based on the information provided, the critical value for a two-tailed test is $t_c=2.009575.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.009575\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{40000-50000}{1000/\sqrt{50}}=-70.711$$

Since it is observed that $|t|=70.711>2.009575=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$  is different than 50000, at the 0.05 significance level.

Using the P-value approach: The p-value $df=49, t=-70.711,$ $two-tailed,$ is $p=0,$ and since $p=0<0.05,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$  is different than 50000, at the 0.05 significance level.