Find the 95% confidence intervals for the variance and standard deviation of the ounces of coffee that a machine dispenses in 12 ounce cups. Assume the variable is normally distributed. The data are given as follows:
12.03 12.10 12.02 11.98 12.00 12.05 11.97 11.99 11.90 12.08
This is what I have, but it does not look correct to me.
µ=12.012, ?^2=0.0034, two critical values are 2.700 and 19.023, using df=9 and 0.025 and 0.975
The 95% confidence interval for variance is
(10-1)(0.0034)/19.023 < ?^2 < (10-1)(0.0034)/2.700 = 0.0016 < ?^2 < 0.1133
The 95% confidence interval for standard deviation is
?0.0016 < ? < ?0.1133 = 0.04 < ? < 0.34
ANY SUGGESTIONS???!!
12.03 12.10 12.02 11.98 12.00 12.05 11.97 11.99 11.90 12.08
This is what I have, but it does not look correct to me.
µ=12.012, ?^2=0.0034, two critical values are 2.700 and 19.023, using df=9 and 0.025 and 0.975
The 95% confidence interval for variance is
(10-1)(0.0034)/19.023 < ?^2 < (10-1)(0.0034)/2.700 = 0.0016 < ?^2 < 0.1133
The 95% confidence interval for standard deviation is
?0.0016 < ? < ?0.1133 = 0.04 < ? < 0.34
ANY SUGGESTIONS???!!