Question

The average length of time for students to enroll has been 50 minutes. If a random sample of 12 students incurred an average enrolment time of 42 minutes with a standard deviation of 11.9 minutes under a new enrolment system, test the hypothesis that the population mean is now less than 50 using a 0.05 level of significance. Assume that the length of time of the population is normal.

Answer 1

Solution.

The null hypothesis H₀: $\mu=50$

The alternative hypothesis H₁: $\mu<50$

Sample size $n=12$

Sample mean $\bar X =42$

Sample standard deviation $\sigma=11.9$

Significance level $\alpha =0.05$

Since the alternative hypothesis is that $\mu$ is Less than this is a one tailed hypothesis test.

Since the sample size is less than 30 and the population standard deviation is not known, we use the T test

$df =n-1=12-1=11$

Since $\alpha = 0.05$ , T $=-1.7959$

$t_c =\dfrac{\bar X -\mu_0 }{S/\sqrt n}=\dfrac {42-50}{11.9/\sqrt12 }$

$t_c =-2.3288$

Since $t_c<T$ we reject the hypothesis.

Since the null hypothesis is rejected, there is sufficient evidence to conclude that the population mean is now less than 50 at 0.05 level of significance.