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 screen 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. Select the relevant null and the alternative hypotheses. b-1. Calculate the value of the test statistic.b-2. Find the p-value.

Answer 1

a)

The null hypothesis is
`H_0: \mu \leq 10`

The alternative hypothesis is
`H_1: \mu > 10`

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

`H_0: \mu \leq 10`

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

`H_1: \mu > 10`

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

`t = (X - \mu)/((s)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that
`\mu = 10`

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.

This means that
`n = 45, X = 10.5, s = 1.8`

Then

`t = (X - \mu)/((s)/(√(n)))`

`t = (10.5 - 10)/((1.8)/(√(45)))`

`t = 1.86`

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.