Normal distribution α(alpha) question

H0: mean = 4000 hours
s =(3,850 - 4,000) / (450 / 25^0.5)
s= -1.6667
from the 2-tail t-table with degree of freedom as 24 (25-1) and 95% confidence level, 2.0639 is obtained.
Since 1.667 is < 2.0639, H0 is accepted.

a level of significance (denoted by a) is chosen. For each level of significance, there corresponds critical value(s), zc. If the value of z calculated from equation (1) is outside the critical values, the result is said to be significant at a level. In this case we reject the null hypothesis. If the value of z is inside the critical values, we do not reject the null hypothesis in favour of the alternative hypothesis and the result is said to be non-significant.

This question need to use t-statistic

H0: The computer will last for 4000 hours

H1: The computer will last for less than 4000 hours

α=0.05, v=25-1=24

critical value: t(0.05,24)=1.711 (I think this question need to use 1-side test)

t(statistic)

=|(3850-4000)/(450/5)|

=1.67

t(statistic)= 1.67&lt;1.711

Conclusion: Don&#39;t reject H0 and conclude that the computer of Laptop computer manufacturer will last for 4000 hours.

α是顯著水準﹐即是型I誤差的可能性﹐要定好α才可以決定個critical value的值

型I誤差: H0是對的﹐但是拒絕H0的概率。