Question

Engineers for a cell phone producer think that using Bluetooth decreases battery life. Average battery life is expected to be 23 hours. A sample of 20 phones was tested for battery life with Bluetooth enabled. The average for the sample was 22.5 hours with a standard deviation of 0.9 hours.

What is the null hypothesis?

What is the alternative hypothesis?

What is the test statistic t ?

What is the t value for a .05 one tailed critical (rejection) region?


Draw the rejection region.

Answer 1

Answer on Question #44304 – Math - Statistics and Probability

Problem.

Engineers for a cell phone producer think that using Bluetooth decreases battery life. Average battery life is expected to be 23 hours. A sample of 20 phones was tested for battery life with Bluetooth enabled. The average for the sample was 22.5 hours with a standard deviation of 0.9 hours.

What is the null hypothesis?

What is the alternative hypothesis?

What is the test statistic $t$ ?

What is the t value for a .05 one tailed critical (rejection) region?

Draw the rejection region.

Solution.

Suppose that, $\mu_0 = 23$ hours, $n = 20$ , $\bar{x} = 22.5$ hours, $s = 0.9$ hours, $\alpha = 0.05$ .

The null hypothesis is $H_0$ : $\mu = \mu_0$ .

The alternative hypothesis is $H_{1}$ : $\mu < \mu_{0}$ (since producer think that using Bluetooth decreases battery life).

The test statistic

$$t = \frac {\bar {x} - \mu_ {0}}{s / \sqrt {n}} = \frac {2 2 . 5 - 2 3}{0 . 9 / \sqrt {2 0}} \approx - 2. 4 8.$$

There are degrees of freedom $n - 1 = 20 - 1 = 19$ . Hence $t$ value for a 0.05 one tailed critical (rejection) region is $t_{\alpha}^{(n-1)} = t_{0.05}^{19} = 1.73$ .

$H_0$ is rejected.

www.AssignmentExpert.com