Need help asap please
1 answer:
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Answer:
Interval [0.67;0.95]
Step-by-step explanation:
Hello!
You need to estimate the population proportion of grain-handling facilities that detect GM beans in shipments with a confidence interval.
The formula you have to use is:
^ρ± * √(^ρ*(1-^ρ)/n)
where ^ρ represents the sample proportion
^ρ= x/n = 17/21 = 0.809 ≅ 0.81
Confidence level 90%
=
= 1.64
Interval
0.81±1.64 * √((0.81*0.19)/21)
0.81±0.140
[0.67;0.95]
With a confidence level of 90% you'd expect that the interval [0.67;0.95] contains the population proportion of grain-handling facilities that detect GM beans in shipments.
Remember:
The confidence level of an interval is the probability under which it is built. This probability indicates that if they build 100 confidence intervals, we expect 90 to contain the value of the population mean we are trying to estimate.
The bigger the confidence level, the bigger is the chance that it contains the estimated parameter, but then again, if you were to study the whole population there is no chance you'll miss the parameter.
The smaller the confidence level, there are fewer chances that it will contain the estimated parameter, but if it does, the estimation would be more accurate than with a bigger level.
You can think it like this, a shooter that hits the target at a distance of 5 meters (low confidence level) is way better than a shooter that hits the target at a distance from 20 cm (high confidence level)
90% Confidence level is acceptable enough to have an accurate estimation of the population parameter.
I hope it helps!