Question

The ABC Company has developed a new cellphone model. The engineering department claims that its battery lasts for 4 days. In order to test this claim, the company selects a random sample of 100 new cellphones so that this sample has a mean battery life of 2.5 days with a standard deviation of 1 day. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=4$

$H_1:\mu\not=4$

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

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=99$ and the critical value for a two-tailed test is $t_c =2.626405.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.626405\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{2.5-4}{1/\sqrt{100}}=-15$$

Since it is observed that $|t|=15>2.626405=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=99$ degrees of freedom, $t=-4$ is $p=0,$ and since $p= 0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 4, at the $\alpha = 0.01$ significance level.