Question

2. A small university is trying to monitor its electricity usage. For a random sample of

Answer 1

Using the t-distribution, it is found that:

a) Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that significantly more electricity is used at the student center, on average, on weekdays than weekend days.

b) The 95% confidence interval for the difference in mean electricity use at the student center between weekdays and weekend days is (3.56, 33.18).

Item a:

At the null hypothesis, it is tested if there is no difference, that is, the subtraction is at most 0, hence:

$H_0: \mu_1 - \mu_2 \leq 0$

At the alternative hypothesis, it is tested if there is difference, that is, the subtraction is positive, hence:

$H_1: \mu_1 - \mu_2 > 0$

The standard errors are given by:

$s_1 = \frac{32.07}{\sqrt{60}} = 4.14$

$s_2 = \frac{43.29}{\sqrt{30}} = 7.9$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_1 - \mu_2 = 112.63 - 94.26 = 18.37$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{4.14^2 + 7.9^2} = 8.92$

The standard deviation are given for the samples, hence, the t-distribution is used.

The test statistic is given by:

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

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

Hence:

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

$t = \frac{18.37 - 0}{8.92}$

$t = 2.06$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 30 + 60 - 2 = 88 df is $t^{\ast} = 1.66$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that significantly more electricity is used at the student center, on average, on weekdays than weekend days.

Item b:

The interval is:

$\overline{x} \pm t^{\ast}s$

Hence:

$\overline{x} - t^{\ast}s = 18.37 - 1.66(8.92) = 3.56$

$\overline{x} + t^{\ast}s = 18.37 + 1.66(8.92) = 33.18$

The 95% confidence interval for the difference in mean electricity use at the student center between weekdays and weekend days is (3.56, 33.18).

A similar problem is given at brainly.com/question/25812826