I get 21.0494777
BUT there is a HUGE flaw in this.
You really need to assume that the pop st deviation is 2.8
and not use LAST year's sample st deviation
The calculation is............
At one office, the average amount of time that workers spend using computers is 21.6 hours with a standard deviation of 2.8 hours. One year later, hypothesis test was done to determine whether the average amount of time spent using computers decreased. For sample of size 70 and significance level of 0.05, find the minimum value of sample mean that makes this test not to reject the null hypothesis.
1)21.05
2)22.15
3)19.55
4)21.60
5)22.50
The answer is 21.05(#1). How do I solve this problem.
Also,
null hypothesis is mu = 21.6
alternative hypothesis is mu < 21.6
Therefore it's a one sided test?