Tomas correctly spelled 18 out of
2 answers:
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Answer:
a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).
b) A sample of 408 is required.
c) A sample of 20465 is required.
Step-by-step explanation:
Question a:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.
This means that
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).
b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?
The margin of error is of:
A sample of n is required, and n is found for M = 0.009. So
Rounding up:
A sample of 408 is required.
c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?
Since we have no estimate, we use
Rounding up:
A sample of 20465 is required.