Answer:
c. Yes because the cut-off is more than the sample mean
Explanation:
The null and alternate hypotheses are
H0 : u= 3.5 ounces vs Ha: u > 3.5 ounces
The mean and standard deviation of the sample are calculated to be 3.76 and 0.24
The test statistics is
t= x`- u/s√n
= 3.76-3.5/0.24/√10
= 0.26/0.07589
=3.426
The critical region for z test is z > ± 2.33
The critical region for t test is t > ± 2.82 for 9 d.f
For both the tests( z and t) the test statistic =3.426 lies in the critical region.
Hence we reject H0 and accept the hypothesis that the new design is better than the old one at a level of significance of 1%.
cut-off = mean + 3 times the standard deviation
= 3.76 + 3* 0.24
cut-off =4.48
Yes, the cut off is more than the sample mean.
So choice c is the best option.