Confidence intervals without the knowing the distribution of the data

Question

I am given the data set containing the values

12.09, 11.18, 9.97, 10.5,0 9.92, 9.97, 11.84, 10.93, 10.70.

I am asked to construct a 90% confidence interval for the mean but I am not told the distribution.

I am told to use the software package R in order to justify any assumptions I make. Even once I have plotted this data in R though I am not seeing a clear distribution and thus cannot construct the confidence interval.

Can anyone help?

I have also found the sample mean to be 10.78889 and the standard deviation to be 0.8029702.

Answer 1

I agree that you do not have enough information to show a clear-cut best way to find the required confidence interval (CI). This is not an uncommon quandary in statistical practice, but rather rare in class exercises. I will indicate three possible methods, each with possible difficulties.

Data and description. There is a stray $0$ in your dataset. Removing it, I get the following in R:

 x = c(12.09, 11.18, 9.97, 10.5,  9.92, 9.97, 11.84, 10.93, 10.70)
 mean(x);  sd(x)
 ## 10.78889
 ## 0.8029702
 length(x)
 ## 9
 shapiro.test(x)

 ##   Shapiro-Wilk normality test

 ## data:  x 
 ## W = 0.9083, p-value = 0.304

 summary(x)
 ## Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 ## 9.92    9.97   10.70   10.79   11.18   12.09 

This agrees with your reported sample mean and variance. Also, the null hypothesis of the Shapiro-Wilk test is that the data are normal, and that hypothesis is not rejected. But with only $n = 9$ observations, it is almost impossible to determine the actual population distribution from which these data arise. A boxplot shows noticeable skewness toward higher values, which might be used to make a (somewhat feeble) argument against a normal population.

Assume normality: t CI. Assuming a normal population, one could find a t confidence interval. These confidence intervals are often remarkably accurate even if the data are not exactly normal. Here are results from R that include a t interval $\bar X \pm 1.859548 S/\sqrt{9},$ where 1.859548 cuts 5% from the upper tail of Student's t distribution with $n - 1 = 8$ degrees of freedom.

 t.test(x, alternative="two.sided", conf.level=0.90)

 ##   One Sample t-test

 ## data:  x 
 ## ... (Irrelevant lines omitted.)
 ## 90 percent confidence interval:    
 ## 10.29117 11.28661 
 ## sample estimates:
 ## mean of x 
 ## 10.78889 

Doing the t interval "by hand" according to the formula above, one obtains the same result:

 pm = c(-1,1)
 mean(x) + pm*qt(.95,8)*sd(x)/sqrt(9)
 ## 10.29117 11.28661

Find CI for median with Wilcoxon procedure. A common nonparametric procedure for finding confidence intervals when the population distribution is unknown is the Wilcoxon procedure, but it finds a confidence interval for the population $median$, and the skewness of the data suggests that the population mean and median may not be the same.

 wilcox.test(x, alternative="two.sided", conf.in=T, conf.level=0.95)

 ##        Wilcoxon signed rank test with continuity correction

 ## data:  x 
 ## ... (Irrelevant lines omitted.)
 ## 95 percent confidence interval:
 ##  9.970033 11.395051 
 ## sample estimates:
 ## (pseudo)median 
 ## 10.74154 

There are some warning messages because there are tied observations in your data (9.97 appears twice) that may interfere with the accuracy of the result. I have not included the warning messages here. It is also worth mentioning that this Wilcoxon procedure (based on 'signed ranks') works best with data from a symmetrical distribution.

Do nonparametric bootstrap with too-small dataset. I don't know if you have studied 'nonparametric bootstrap confidence intervals'. If so, that might be a better way to find a confidence interval for the population mean. The nonparametric bootstrap procedure does not make any assumption about the population distribution. The disadvantage to doing a bootstrap CI is that the sample size is smaller than recommended for that procedure. That procedure is a little difficult to explain from scratch if you haven't studied it. (You would need a 'bias corrected' version on account of the skewness of the data.)

If you have studied boostrapping, then I suppose that may be what the question intended. Try it. If you need help, please edit the request for a bootstrap CI into your Question, and I (or someone else) may provide the details. Also, please address a Comment to me below.

Addendum (posted later for completeness): A bootstrap "90% CI" is $(10.4, 11.2)$. The true coverage probability of such intervals based on small samples is typically a little less than 90%. However, for practical purposes, this interval is not much different from the t interval above.

Conclusion. In summary, you are not left with a clear-cut course of action: For the t CI you need to make the leap of assuming nearly-normal data, for a Wilcox CI you need to assume that population mean and median are equal (and wonder about inaccuracy due to ties), and for a bootstrap CI you need to wonder if you have a large enough dataset. However, notice that the CIs produced by the three methods are not much different.

Answer 2

To construct a confidence interval, you need the mean and the standard deviation. Would you be able to estimate the mean and the standard deviation from the underlying population using the sample you were given?