Someone help,please.I don't understand very well.
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a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).
b) The margin of error is: 2.45%.
c) The 90% confidence is the level of confidence that the true population percentage is in the interval.
d) The needed sample size is: 271.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions has the bounds given by the rule presented as follows:
The margin of error is given by:
The variables are listed as follows:
is the sample proportion, which is also the estimate of the parameter.
- z is the critical value.
- n is the sample size.
The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of , so the critical value is z = 1.645.
The sample size and the estimate are given as follows:
The margin of error is of:
The interval is given by the estimate plus/minus the margin of error, hence:
- The lower bound is: 10 - 2.45 = 7.55%.
- The upper bound is: 10 + 2.45 = 12.45%.
For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:
n = 271 (rounded up).
More can be learned about the z-distribution at brainly.com/question/25890103
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