Question

An online streaming service providing television programs claims that a 30-minute program will stream with advertisements that average 45 seconds. A consumer advocacy group is investigating to see if this claim is true. They recorded the times of 21 randomly selected advertisements. The times are listed below:
Time (seconds) 45 45 43 50 50 45 43 50 45 49 46 48 42 46 44 52 48 45 46 50 48
The mean and standard deviation for these times are 46.67 and 2.78 seconds respectively. Do these data provide convincing evidence that the true mean advertisement length is longer than 45 seconds?

Answer 1

Answer:

The p-value is significantly low(<0.01), which means that the data provides convincing evidence that the true mean advertisement length is longer than 45 seconds.

Step-by-step explanation:

An online streaming service providing television programs claims that a 30-minute program will stream with advertisements that average 45 seconds. Test if the true mean advertisement length is longer than 45 seconds.

At the null hypothesis, we test if the mean time is of 45 seconds, that is:

$H_0: \mu = 45$

At the alternate hypothesis, we test if the mean is more than 45 seconds, that is:

$H_1: \mu > 45$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

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.

45 is tested at the null hypothesis:

This means that $\mu = 45$

They recorded the times of 21 randomly selected advertisements. The mean and standard deviation for these times are 46.67 and 2.78.

This means that $n = 21, X = 46.67, s = 2.78$

Value of the test-statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{46.67 - 45}{\frac{2.78}{\sqrt{21}}}$

$t = 2.75$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 46.67, which is a right-tailed test, with t = 2.75 and 21 - 1 = 20 degrees of freedom.

With the help of a calculator, this p-value is of 0.0062.

The p-value is significantly low(<0.01), which means that the data provides convincing evidence that the true mean advertisement length is longer than 45 seconds.