URGENT: Can someone help me solve these problems?
1 answer:
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Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
Step-by-step explanation:
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 .
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that
80% 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 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
To find the reciprocal of a number or fraction, just flip the numerator and denominator. Whole numbers are always over 1.
2 =
The reciprocal of
or .5 (one-half)
The reciprocal of
2 =
The reciprocal of
The reciprocal of