Question

We claim that the average weight of our "product" is 50 pounds, with a standard deviation of 2 pounds. We take a sample of 50 units, with a mean of 49.95 pounds and a standard deviation of 1.9999 pounds. What is a 95% prediction interval for the mean weight of the NEXT unit of production from our process? Use Z for ease of calculation.

Answer 1

Answer:

49.95+/-0.5543

= ( 49.3957, 50.5043) pounds

the 95% confidence interval (a,b) = ( 49.3957, 50.5043) pounds

And to 2 decimal points;

the 95% confidence interval (a,b) = ( 49.40, 50.50) pounds

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 49.95 pounds

Standard deviation r = 1.9999 pounds

Number of samples n = 50

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

49.95+/-1.96(1.9999/√50)

49.95+/-1.96(0.282828570338)

49.95+/-0.554343997864

49.95+/-0.5543

= ( 49.3957, 50.5043) pounds

Therefore, the 95% confidence interval (a,b) = ( 49.3957, 50.5043) pounds