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Answer:
represent the proportion estimated of subjects reported with dizziness
n = 268 represent the random sample selected
represent the proportion of subjects No reported with dizziness
E= 0.04 = 4% represent the margin of error for the confidence interval
and this value represent the significance level of the test or the probability of error type I
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
Solution to the problem
The confidence interval for the mean is given by the following formula:
The margin of error is given by:
And for this case we have the following info:
represent the proportion estimated of subjects reported with dizziness
n = 268 represent the random sample selected
represent the proportion of subjects No reported with dizziness
E= 0.04 = 4% represent the margin of error for the confidence interval
and this value represent the significance level of the test or the probability of error type I
The only solution is (-3/2, 4)
read my attachment
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1, when you factor you get 6x(x^2-1)+4(x^2-1)=