Answer:
1.) Yes,
2.) (10.0017518 ; 10.0022482)
Explanation:
1.) Since the average is > mean, then it shows that scale isn't well calibrated.
2.)
Sample mean = 10.002
Standard Error = σ/√n = 0.0002/√5
df = n - 1 = 5 - 1 = 4
Tcritical(0.05, 4) = 2.776
Margin of Error = Tcritical * standard error
Margin of Error = 2.776*0.0002/√5 = 0.0002482
Confidence interval :
Mean ± margin of error
Lower boundary = 10.002 - 0.0002482 = 10.0017518
Upper boundary = 10.002 + 0.0002482 = 10.0022482
(10.0017518 ; 10.0022482)
C.)
Since 10 does not fall within the interval ; then we can agree with our assertion in (1).
Confidence interval gives a range of of all possible values and not just a single point value.