To assess the accuracy of a laboratory scale, a reference weight known to weigh exactly 10 grams (g) is weighed repeatedly. The scale readings are Normally distributed with standard deviation sigma = 0.0002 g. The reference weight is weighed 5 times on that scale. The mean result is 10.002} g. a) Do the 5 weighings give good evidence that the scale is not well calibrated (that is. its mean mu for weighing this weight it not 10 g)?
b) Give a 95% confidence interval for the mean weight on this scale for all possible measurements of the reference weight. What do you conclude about the calibration of this scale?
c) Compare your results in pans (a) and (b). Explain why the confidence interval is more informative than the test result.

1.) Yes,

2.) (10.0017518 ; 10.0022482)

Step-by-step 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.

he sum of the measures of two complementary angles is 90 degrees

Step-by-step explanation:

we know that

If two angles are complementary, then their sum is equal to 90 degrees

In this problem we have that

m∠ABD and m∠DBC are complementary

so

m∠ABD + m∠DBC=90° > by complementary angles

and

(m∠ABD + m∠DBC)+m∠EBC=180° > by a linear pair

Find the measure of angle EBC

substitute the given values

(90°)+m∠EBC=180°

∠EBC=180°-90°=90°