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.


Do you reject or fail to reject the null hypothesis? Show why you made your choice.


What is the approximate p value for the test?

Answer 1

Answer on Question #44387 – 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.

Do you reject or fail to reject the null hypothesis? Show why you made your choice.

What is the approximate p value for the test?

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 a critical $t$ 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 at $\alpha = 0.05$ , because the test statistics value $(-2.48)$ falls into the rejection region $(t \leq -1.73)$ .

The approximate p-value for the test is

$p = T_{n - 1}(|t|) = T_{19}(2.48) = TDIST(2,48;19;1) = 0.0113$ (via Excel function TDIST).

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