Answer to Question #118010 in Statistics and Probability for Cynthia De Vera

Question

The average length of time for students to register in the 1st semester at a certain school has been 50mins. A new registration procedure is being tested. If random samples of 20 students have an average of 45mins with a standard deviation of 9mins under the system, can you conclude that the new system is faster than the old? Use 0.01 level of significance.

Answer 1

$a_0=50\\ n=20\\ \overline{x}=45\\ s=9\\ \alpha=0.01\\ H_0: \mu=a_0=50, H_1: \mu<a_0=50\\ \mu\text{ --- the average length of time for students}\\ \text{to register under the new system}.\\ \text{We assume that the length of time for students}\\ \text{to register has normal distribution}.\\ \text{We will use the following random variable as a criterion:}\\ T=\frac{(\overline{X}-a_0)\sqrt{n}}{s}\\ T\text{ has t-distribution with } k=n−1 \text{ degrees of freedom.}\\ \text{Observed value:}\\ t_{obs}=\frac{(45-50)\sqrt{20}}{9}\approx -2.48\\ \text{Critical value (one-sided):}\\ t_{cr}=t_{cr}(\alpha;k)=t_{cr}(0.01;19)\approx 2.54\\ (-\infty, -2.54)\text{ --- critical region}\\ t_{obs}\text{ does not fall into the critical region. So we accept } H_0.\\ \text{We cannot conclude that the new system is faster than the old}.$