Question

DIRECTIONS: In each problem below, give the null and alternative hypothesis and

identify whether it is right-tailed, left-tailed or two-tailed test.

3. A quality control engineer is testing the battery life of a new smartphone. The company

is advertising that the battery lasts 24 hours on full- charge, but the engineer suspects that the

battery life is actually less than that. They take a random sample of 50 of these if their average

battery life is significantly less than 34 hours.

4. In the past, the mean running time for a certain type of radio battery has been 9.6 hours.

The manufacturer has introduced a change in then production method and wants to perform a

hypothesis test to determine whether the mean running time has changed as a result.

5. In a random sample of 400 electronic gadgets, 14 were found to be defective. The

manufacturer wants to claim that more than 5% of all the gadgets are defective. Test this claim at

the 0.01 level of significance.

Answer 1

3. The null hypothesis states that the battery life of a new smartphone will be greater than or equal to 24 hours.

The alternative hypothesis states that the battery life of a new smartphone will be less than 24 hours.

$H_0: \mu\ge24$

$H_1: \mu<24$

B. This corresponds to a left-tailed (directional, one-tailed) test.

4. The null hypothesis states that the mean running time for a certain type of radio battery will be equal to 9.6 hours.

The alternative hypothesis states that the mean running time for a certain type of radio battery will not be equal to 9.6 hours.

$H_0: \mu=9.6$

$H_1: \mu\not=9.6$

B. This corresponds to a two-tailed test.

5. The following null and alternative hypotheses need to be tested:

$H_0:p\le0.05$

$H_a:p>0.05$

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z: z > 2.3263\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.035-0.05}{\sqrt{\dfrac{0.05(1-0.05)}{400}}}=-1.376$$

Since it is observed that $z=-1.376<2.3263=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, is $p =P(Z>-1.376)=0.915589,$ and since $p= 0.915589>0.01=\alpha,$ it is concluded that the null hypothesis is notrejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is more than 5%, at the $\alpha = 0.01$ significance level.