Answer:
Option B is right
Step-by-step explanation:
[tex]H_0: x bar =2\\H_a: x bar <2[/tex]
(One tailed test at 5% significance level)
n =16 and x bar =1.97
s =0.1
Std error of mean = [tex]\frac{s}{\sqrt{n} } \\=0.025[/tex]
Mean diff = [tex]1.97-2 =-0.03[/tex]
t statistic =Mean diff/se =-1.2
df =16-1=15
p value =0.124375
(B) do not reject the null because the test statistic (-1.2) is > the critical value (-1.7531).