Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 70 cookies. The mean is 22.24 and the standard deviation is 2.44. .Construct a 98​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies. How do I find the degree of freedom, which is 69, using the chi critical value table? 69 is not listed on there, just 60 and 70. Not sure how to use my TI-84 calculator to figure it out.

Answer 1

Given :

• There is a sample of 70 cookies i.e. ,  n = 70
• Mean is 22.4 i.e , = 22.4
• Standard Deviation is 2.44 i.e. ,   = 2.44

We want a 98 % Confidence Interval for the Population Standard Deviation (   ) .

The Test - Statistic is as follows :

98 % Confidence Interval implies :

For this , We require the   values with 69 degrees of freedom at 99 % and 1 % percentage points .

Using distribution tables ,

We have values given at 60 and 70 degrees of freedom . So , we get the value at 69 degrees of freedom , we use linear interpolation which implies :

................. where , is the required value at given value of x .

• 
• 

............................................

Similarly :

• 
• 

......................................

Therefore :

  with 98 % Confidence .