The average length of time for student to register summer classes at a certain university has been 50 minutes with a standard deviation of 10 minutes. A new registration procedure using modern computing machine is being tried. If a random sample of 12 students had a n average registration of 42 minutes with a standard deviation of 11.9 minutes under the new system, test the hypothesis that the population mean is now less than 50 minutes. (critical value is +/- 1.796)
The following null and alternative hypotheses need to be tested:
"H_0:\\mu\\ge50"
"H_1:\\mu<50"
This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.
Based on the information provided, the significance level is "\\alpha = 0.05," "df=n-1=11" degrees of freedom, and the critical value for a left-tailed test is "t_c = -1.796."
The rejection region for this left-tailed test is "R = \\{t:t<-1.796\\}."
The t-statistic is computed as follows:
Since it is observed that "t=-2.3288<-1.796=t_c," it is then concluded that the null hypothesis is rejected.
Using the P-value approach:
The p-value for left-tailed, "df=11" degrees of freedom, "t=-2.3288," is "p= 0.019977\n," and since "p= 0.019977\n<0.05=\\alpha," it is concluded that the null hypothesis is rejected.
Therefore, there is enough evidence to claim that the population mean "\\mu"
is less than 50, at the "\\alpha = 0.05" significance level.
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