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I've been asked a few questions in regards to confidence interval width estimates and cant seem to find the answer on a practice quiz i'm doing.

You have a sample with sample mean x̄=100 and sample standard deviation s=10 . The distribution of the sample means of the bootstrap samples is bell-shaped with the standard error SE=2 . What is the width of a 95% confidence interval for the population mean?

I calculated a 95% confidence interval with x̄ ± 2 ⋅ SE = 4 and then multiplied that by 2 to get the width of the confidence interval which was equal to 8. however the answer is wrong and after going through my notes I cant find where I have gone wrong.

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  • $\begingroup$ were there options to choose from? $\endgroup$ – Saketh Malyala Apr 30 at 1:54
  • $\begingroup$ your answer looks correct, because 95% confidence = 1.96 standard deviations in either direction. $\endgroup$ – Saketh Malyala Apr 30 at 1:54
  • $\begingroup$ is 7.84 an answer choice? $\endgroup$ – Saketh Malyala Apr 30 at 1:55
  • $\begingroup$ it wasn't multiple choice question but 8 is considered incorrect, I will try 7.84 next time the question pops up $\endgroup$ – lowy95 Apr 30 at 2:02
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Your sample size is $n=25$ because, $s/SE= 5 = \sqrt{n}$. Thus, the variance of the population is $n/(n-1)s^2 = 10.42^2$. Hence the SD of the distribution of the mean is $10.42/5 = 2.08$ and the $95\%$ conf. int. for the population mean is approximately $4 \times 2.08 = 8.32 $.

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  • $\begingroup$ @Henry -- absolutely so -- I somehow misread "sample standard deviation s=10" to mean something else.... I fixed the answer. $\endgroup$ – dnqxt Apr 30 at 23:29

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