These are my last questions plz help me :)
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Answer:
lower limit =0.261
upper limit =0.369
Step-by-step explanation:
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".
Description in words of the parameter p
represent the real population proportion of people that have experienced bullying
X= 63 people in the random sample that have experienced bullying
n=200 is the sample size required
represent the estimated proportion of people that have experienced bullying
represent the critical value for the margin of error
The population proportion have the following distribution
Numerical estimate for p
In order to estimate a proportion we use this formula:
where X represent the number of people with a characteristic and n the total sample size selected.
represent the estimated proportion of people that they were planning to pursue a graduate degree
Confidence interval
The confidence interval for a proportion is given by this formula
For the 90% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 90% confidence interval would be given (0.261;0.369).
We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).
lower limit =0.261
upper limit =0.369