1. Unit 7 Section 7-5

2. 7-5: Confidence Intervals for Variances and Standard Deviations  In 7-2 to 7-4 we determined confidence intervals for means and proportions. Confidence intervals can also be determined for standard deviations and variance.  For example : If we were manufacturing pipes, you would want to keep the variation of diameters down in order for the pieces to fit together and avoid being scrapped.  In order to find a confidence interval for standard deviation and variance, we use a new distribution known as a chi-square distribution.

3.  A chi-square distribution is family of curves like the t distribution (based on degrees of freedom).  Chi-square variable can not be negative.  The chi-square distribution is positively skewed  The area under a chi-square distribution is 1 or 100%  Values for variance and standard deviation are assumed to be normally distributed.  Two sets of values will be used in the formula for chi- square (left and right side of the table). Section 7-5

4. Finding the left and right chi-squared value First determine what confidence interval we wish to find (in decimal form). Subtract the percentage from 1. Divide this value by two to determine value of X 2 right. Subtract the value of X 2 right from 1 to get X 2 left.

5.  Example 1: Find the values for X 2 right and X 2 left for a 90% confidence interval when n = 25. Section 7-5

6. Formula for Confidence Interval for a Variance  where  Rounding Rule : Round to one more decimal place than the original data (or same as sample standard deviation). Section 7-5

7. Formula for Confidence Interval for a Standard Deviation  where  Note : if the problem gives sample variance you do not need to square it! (remember, s 2 is sample variance) Section 7-5

8.  Example 2: Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a sample of 20 cigarettes has a standard deviation of 1.6 milligrams. Section 7-5

9.  Example 3: Find the 90% confidence interval for the variance and standard deviation for the price in dollars of an adult single-day ski lift ticket. The data represents a selected sample of nationwide ski resorts. Assume the variable is normally distributed. 5954535251 3949464948 Section 7-5

10. Homework:  Pg 381 : #’s 1 - 7 Section 7-5