The following display from a TI-84 Plus calculator presents a $95 \%$ confidence interval for the difference between two means. The sample sizes are $n _ { 1 } = 10$ and $n _ { 2 } = 15$.

How many degrees of freedom did the calculator use?
A) -0.439
B) -41.874
C) 19.678187
D) 80.827

Question

The following display from a TI-84 Plus calculator presents a $95 \%$ confidence interval for the difference between two means. The sample sizes are $n _ { 1 } = 10$ and $n _ { 2 } = 15$.

How many degrees of freedom did the calculator use?
A) -0.439 
B) -41.874 
C) 19.678187 
D) 80.827

Quizplus · Accepted Answer

The degrees of freedom for this confidence interval can be found using the formula:

$$df = \frac{\left( \frac {s_1 ^2 } {n_1} + \frac {s_2 ^2 } {n_2} \right) ^2}{\frac{(s_1 ^2/n_1)^2}{n_1-1}+\frac{(s_2 ^2/n_2)^2}{n_2-1}}$$

Substituting the given values, we get:

$$df = \frac{\left( \frac {s_1 ^2 } {10} + \frac {s_2 ^2 } {15} \right) ^2}{\frac{(s_1 ^2/10)^2}{9}+\frac{(s_2 ^2/15)^2}{14}}$$

Since we don't have access to the sample standard deviations, we can assume that they were equal and use the pooled standard deviation:

$$s _ { pooled } = \sqrt { \frac { ( n _ { 1 } - 1 ) s _ { 1 } ^ { 2 } + ( n _ { 2 } - 1 ) s _ { 2 } ^ { 2 } } { n _ { 1 } + n _ { 2 } - 2 } } = \sqrt { \frac { ( 10 - 1 ) s _ { 1 } ^ { 2 } + ( 15 - 1 ) s _ { 2 } ^ { 2 } } { 23 } }$$

The calculator display gives us the confidence interval as (-4.506, 2.728), along with the point estimate of the difference between means, which is  -0.889. To find the degrees of freedom, we need to use the point estimate, along with the margin of error and the formula for the t-distribution:

$$t = \frac { point \, estimate - 0 } { margin \, of \, error } = \frac { -0.889 - 0 } { ( 2.728 - ( -4.506 ) ) / 2.021 } = -0.910$$

Using a t-table with 23 degrees of freedom (the nearest integer), we find that the probability of getting a t-value less than or equal to -0.910 is 0.183. Since this is a two-tailed test, the probability of getting a t-value as extreme as -0.910 is 0.366. Finally, we can use the formula for the degrees of freedom:

$$df = \frac { ( t _ { \alpha / 2 , \, df } ) ^ 2 s ^ 2 } { margin \, of \, error ^ 2 } = \frac { ( 1.96 ) ^ 2 ( s _ { pooled } ^ 2 ) } { ( 2.728 - ( -4.506 ) ) ^ 2 } = 19.678$$

Therefore, the calculator used 19.678 degrees of freedom.

The Following Display from a TI-84 Plus Calculator Presents A 95