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Is this possible to compute?

From a normal distribution we took a sample $( 11, 13, 13, 14, 14, 14, 15, 15, 17, 18).$ Compute the $98 \%$ confidence interval of the expected value.

I'm studying statistics on my own but I don't know what formula should I use here.

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  • $\begingroup$ Can you parametrize the stated normal distribution in terms of this expected value? $\endgroup$
    – Michael M
    Commented Dec 2, 2013 at 20:09
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    $\begingroup$ Most samples from normal distributions do not give integers only $\endgroup$
    – Henry
    Commented Dec 2, 2013 at 20:35
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    $\begingroup$ I think that it is a textbook question, which sometimes gives this type of information claiming it to be normal, especially if it is a social science statistics book... $\endgroup$
    – tomka
    Commented Dec 2, 2013 at 20:40
  • $\begingroup$ @tomka in spite of such assertions by some textbooks, a collection of small integer outcomes aren't from a normal distribution, except as a very rough approximation to the actual cdf. $\endgroup$
    – Glen_b
    Commented Dec 2, 2013 at 22:53
  • $\begingroup$ I wasn't claiming that it is OK to do so, I was explaining why/how the OP might have come up with this flawed exercise. In fact I was sarcstic. $\endgroup$
    – tomka
    Commented Dec 2, 2013 at 23:12

1 Answer 1

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You could add the tag self-study to your question.

First estimate the expectation of the normal distribution by the sample mean, then the sample variance, and the square root of the sample variance. 

> x<-c(11,13,13,14,14,14,15,15,17,18)
> mean(x); var(x)
[1] 14.4
[1] 4.044444
> sqrt(var(x))
[1] 2.01108

The confidence interval is contructed from this information. Hint: its upper limit is mean(x)+qt(0.99,df=n-1)*sqrt(var(x))/sqrt(n).

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    $\begingroup$ The OP may never have seen R code. You'd be better placed to give relevant formulae first, and then at the very least explain that you're using R. Even if the OP has seen it, some later readers won't have. $\endgroup$
    – Glen_b
    Commented Dec 2, 2013 at 22:55
  • $\begingroup$ So I got x<-c(11,13,13,14,14,14,15,15,17,18) mean(x)+qt(0.99,df=10-1)*sqrt(var(x))/sqrt(10) mean(x)-qt(0.99,df=10-1)*sqrt(var(x))/sqrt(10) [1] 16.19432 [1] 12.60568 . Is this correct? $\endgroup$
    – guest
    Commented Dec 3, 2013 at 12:31
  • $\begingroup$ This looks good! $\endgroup$
    – tomka
    Commented Dec 3, 2013 at 12:36

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