K&M has contracted with various companies that make these drill bits to supply 5
ID: 3126193 • Letter: K
Question
K&M has contracted with various companies that make these drill bits to supply 50-pound buckets of the powder to them on a regular schedule. Therefore, K&M wants to make sure that it is actually, on average, providing the 50-pound amount of powder that it has agreed to provide in the contract.
The quality control supervisor at K&M decides to investigate the process by calculating a confidence interval for the true mean weight of the buckets it sells that are labeled “50 pounds.” Taking a random sample of 30 buckets, the supervisor finds that the sample mean weight was 49.7 pounds, with a standard deviation of 0.30 pounds.
a). The supervisor will calculate either a 90%, 95%, or 99% confidence interval. Suppose that, among these three levels, the supervisor wants to use the level of confidence that will give him the narrowest interval width. Using that level of confidence, calculate the interval. Calculate and report only one (1) interval for this problem. Show your work; an answer, even if “correct,” with no work shown is worth 0 points.
b). Referring to part (a), now suppose the supervisor wants to use the level of confidence among those three that will give him the most assurance of being “right” about the true mean of the process. Using that level of confidence, calculate the interval. Again, calculate and report only one (1) interval for this problem. Show your work; an answer, even if “correct,” with no work shown is worth 0 points.
c). Interpret the interval you calculated in (b) in the context of the problem. That is, your answer must use the words of the problem, not general terms like "population mean": the population mean of what? Be specific in your answer.
d). Can the supervisor reasonably conclude that the company is really supplying 50-pound buckets to its customers? Why or why not?
Explanation / Answer
a. With a higher confidence 90% will get narrower interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=49.7
Standard deviation( sd )=0.3
Sample Size(n)=30
Confidence Interval = [ 49.7 ± Z a/2 ( 0.3/ Sqrt ( 30) ) ]
= [ 49.7 - 1.64 * (0.055) , 49.7 + 1.64 * (0.055) ]
= [ 49.61,49.79 ]
b.
We calculate the interval at 99% confidence interval, that will give him the
most assurance of being “right”
Confidence Interval = [ 49.7 ± Z a/2 ( 0.3/ Sqrt ( 30) ) ]
= [ 49.7 - 2.58 * (0.055) , 49.7 + 2.58 * (0.055) ]
= [ 49.559,49.841 ]
c.
Interpretations:
1) We are 95% sure that the interval [ 49.559,49.841 ] contains the true population mean
d)
No, it is out of bound in interval
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.