6 squirrels were found to have an average weight of 9.2 ounces with a sample standard deviation is 0.7. Find the 95% confidence interval of the true mean weight.
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Ignoring for the moment the small sample size, we can find the z score for the true mean weight: z=(9.2-m)/0.7, where m is the true mean weight and 0.7 is the given sample standard deviation. From the normal distribution table a 95% probability corresponds to a z score of 1.645, so 1.645=(9.2-m)/0.7; 1.1515=9.2-m, so m=9.2-1.1515=8.05 approx.

However, properly applying the confidence level of 95%, a=1-0.95=0.05 and the critical probability is 1-a/2=0.975. The number of degrees of freedom is 6-1=5, and the standard error is 0.7/sqrt(6)=0.28577. We can work out the critical value from t score tables: 2.571, given df=5 and the critical probability of 0.975. The margin of error is therefore 2.571*0.28577=0.735 and the true mean weight is 9.2+0.74 approx. This range puts m as calculated above out of range (9.2-0.74=8.46). The range for the true mean weight is 8.46 to 9.94.

 

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