In an electronic laboratory we want to estimate the time it takes to assemble a
ID: 3208873 • Letter: I
Question
In an electronic laboratory we want to estimate the time it takes to assemble a particular data component. To estimate the expected time this work takes, , we have registered the assembly time, X, to 40 technicians. The result gives an average (X with a line over) of 17.43 minutes and a standard deviation in measurements, S = 1.98 minutes.
a) Find a 95% confidence interval for the expected time it takes to assemble this data component.
b) The results will now be presented as a measurement result given by the average of the measurements and standard uncertainty based on 40 measurements. What coverage factor (integer 1) should we choose if we want the result to correspond to 95% confidence intervall?
c) What confidence interval (80%, 90% and 95%) will be such that we can say that a measure consisting of observed average, does not deviate more than 30 seconds from the expected time?
Explanation / Answer
Answer to part a)
we got sample size n = 40
Sample mean (x bar) = 17.43
Sample standard deviation s = 1.98
.
The formula of confidence interval is:
x bar - t*s/root (n) , xbar + t*s/root(n)
.
The T critical value for df = 40-1 = 39 and confidence level 0.95 can be obtained from the T table
We t = 0.063112
[I got the exact value with the help of excel command =T.INV.2T(0.95,39)
.
On plugging the values we get:
17.43 - 0.063112 * 1.98/root(40) , 17.43 + 0.063112 * 1.98/root(40)
17.41024 , 17.44976
Thus the 95% confidence interval is 17.41024 , 17.44976
.
Answer to part b)
for 95% confidence interval we generally make use of the coverage factor 2
Thisimplies 95% of the samples contain the population mean within 17.43 -2*1.98 and 17.43+2*1.98
.
Answer to part c) In order to find that the observed average is not more than 30 secs. This means the margin of error is assumed to be 0.5
The T critical value for 80% confidence interval is 0.255082
This interval provides us the deviation within 30 seconds
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