Question

a company has two offices with 50 employees each. The employees in each office report on their daily commute time the results are shown in the table. Is there a significant difference between the commute times for the two populations? Use a 0.05 significance level.​

Answer 1

Step-by-step explanation:

We want to compare the means, so find the difference and the corresponding standard deviation:

μ = μ₁ − μ₂

μ = 25 − 23

μ = 2

σ = √(σ₁²/n₁ + σ₂²/n)

σ = √(3.1²/50 + 3.2²/50)

σ = 0.63

Using a 95% confidence level, z = 1.96.  So the interval is:

CI = 2 ± 1.96 × 0.63

CI = 2 ± 1.2

There is a 95% probability that the true difference in means is between 0.8 and 3.2.  Since 0 is not in this interval, we can conclude that there is a significant difference between the commute times for the two populations.

Answer 2

Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value for the two-tailed test, it is found that there is enough evidence to conclude that there is a significant difference between the commute times for the two populations.

What are the hypothesis tested?

At the null hypothesis, it is tested if there is no difference, that is:

$H_0: \mu_A - \mu_B = 0$

At the alternative hypothesis, it is tested if there is a difference, that is:

$H_1: \mu_A - \mu_B \neq 0$.

What are the mean and the standard error for the distribution of differences?

For each sample, the mean and the standard error are given by:

$\mu_A = 25, s_A = \frac{3.1}{\sqrt{50}} = 0.4384$

$\mu_B = 23, s_B = \frac{3.2}{\sqrt{50}} = 0.4525$

For the distribution of differences, we have that:

$\overline{x} = \mu_A - \mu_B = 25 - 23 = 2$

$s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.4384^2 + 0.4525^2} = 0.63$

What is the test statistic?

It is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{2 - 0}{0.63}$

$t = 3.17$

What is the decision?

The critical value for a two-tailed test, as we are testing if the mean is different of a value, with 50 + 50 - 2 = 98 df and a significance level of 0.05 is given by $|t^{\ast}| = 1.9845$

Since the absolute value of the test statistic is greater than the critical value for the two-tailed test, it is found that there is enough evidence to conclude that there is a significant difference between the commute times for the two populations.

More can be learned about the t-distribution at https://brainly.com/question/16162795