Question 1196689: A hardware manufacturer produces bolts used to assemble various machines. Assume that the diameter of bolts produced by this manufacturer has an unknown population mean, μ, and the standard deviation is 0.1 mm. Suppose the average diameter of a simple random sample of 50 bolts is 5.11 mm.
What is the width of a 95% confidence interval for μ?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Given info:
n = 50 = sample size
xbar = 5.11 = sample mean
sigma = 0.1 = population standard deviation
Side note: In many real world situations, we often won't know what sigma is because it's a population parameter. The goal of statistics is to estimate such parameters.
At 95% confidence, the z critical value is roughly z = 1.96
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 95% confidence level.
Use those values to find the following.
E = margin of error
E = z*sigma/sqrt(n)
E = 1.96*0.1/sqrt(50)
E = 0.02771858582251
E = 0.027719
The result is approximate
The value of E is the distance from the center to either endpoint.
Think of it as the radius of interval.
It's half the confidence interval width, so the full width is 2*E = 2*0.027719 = 0.055438
You could also follow these steps
L = lower bound
L = xbar - E
L = 5.11 - 0.027719
L = 5.082281
U = upper bound
U = xbar + E
U = 5.11 + 0.027719
U = 5.137719
The width of the confidence interval is U - L = 5.137719 - 5.082281 = 0.055438 which is the distance from the lower bound L to the upper bound U.
These steps are handy if you are given the confidence interval in the format (L,U) or the format L < mu < U.
Note how U - L = (xbar+E)-(xbar-E) = 2E, so the xbars cancel out and the E's double up.
Answer: 0.055438 approximately
Round that value however needed.
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