Question

A phone manufacturer wants to compete in the touch screen phone market. He

understands that the lead product has a battery life of just 5 hours. The

manufacturer claims that while the new touch phone is more expensive, its battery

life is more than twice as long as that of the leading product. In order to test the

claim, a researcher samples 45 units of the new phone and finds that the sample

battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.


a. Set up the relevant null and alternative hypotheses.

b. Compute the value of the appropriate test statistic.
c. Use the critical value approach to test the phone manufacturer’s claim at

alpha= 0.05.

plz answer​

Answer 1

Answer:

since the test statistic z=1.863389981 is greater than the critical value (1.6449), we reject Ho at the 5% level of confidence and conclude that the battery life of the new battery is not more than 10 hours.

Step-by-step explanation:

a)

The claim is that the battery life of the new battery is more than twice as long as that of the leading product.

Therefore;

Null hypothesis- battery life of the new battery is more than 10 hours

Alternative hypothesis- the battery life of the new battery is not more than 10 hours

ie

$H_o: \mu>10\\H_a:\mu\leq 10$

b)

Since n=45 is large enough, we can approximate the population to be normally distributed.

The test statistic is ;

$z=\frac{\bar x-\mu_o}{\frac{\sigma}{\sqrt{n} } }\\ \\z=\frac{10.5-10}{\frac{1.8}{\sqrt{45} } }\\ \\z=1.863389981$

z=1.863389981

c) at $\alpha$=0.05,

the critical value for the single sided z-test is  1.6449

since the test statistic z=1.863389981 is greater than the critical value (1.6449), we reject Ho at the 5% level of confidence and conclude that the battery life of the new battery is not more than 10 hours.