Answer:
a) For this condition (α=0.05, df=64, one-sided test), the critical value is t=1.669.
b) The P-value is 0.46
c) The null hypothesis can not be rejected
Step-by-step explanation:
The null and alternative hypothesis are:
[tex]H_0: \mu \leq 1.1 \\\\H_a: \mu > 1.1[/tex]
The significance level is α=0.05.
The degrees of freedom are 64.
The standard deviation of the sample is 0.02 mm.
For this condition (α=0.05, df=64, one-sided test), the critical value is t=1.669.
The test statistic t is calculated as:
[tex]t=\frac{M-\mu}{s_m}=\frac{1.102-1.100}{0.020} =\frac{0.002}{0.020}=0.1[/tex]
As t=0.1 belong to the acceptance region, the null hypothesis can't be rejected.
The P-value is 0.46, and it is bigger than the level of significance. The conclusion is the same: the null hypothesis can not be rejected as there is no strong evidence against it.
