Answer to Question #348316 in Statistics and Probability for Dawit

Question

A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, e-mails, impromptu visits, etc.). A random sample of 50 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the population standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to the results of the study.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=126$

$H_1:\mu\not=126$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=49$ and the critical value for a two-tailed test is $t_c =1.676551.$

The rejection region for this two-tailed test is $R = \{t:|t|>1.676551\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{108-126}{20/\sqrt{50}}=-6.364$

Since it is observed that $|t|=6.364>1.676551=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=49$ degrees of freedom, $t=-6.364$ is $p=0,$ and since $p= 0<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 126 minutes=2.1hours, at the $\alpha = 0.10$ significance level.

The corresponding confidence interval is computed as shown below:

$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}-t_c\times\dfrac{s}{\sqrt{n}})$

$=(108-1.676551\times\dfrac{20}{\sqrt{50}},108+1.676551\times\dfrac{20}{\sqrt{50}})$

$=(103.258, 112.742)$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $103.258 < \mu < 112.742,$ which indicates that we are 90% confident that the true population mean $\mu$  is contained by the interval $(103.258, 112.742).$