Constructing a Confidence Interval for the Slope of a Regression Line, problem 3

In this video, we demonstrate how to construct a confidence interval for the slope of a regression line. In this problem, most of the needed values are provided, except the critical value which comes from a student t distribution with n -2 degrees of freedom.
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[@blew319]

Great video. I have been stumped on the confidence interval of the slope but now I at least get the formula. If you don’t already have a video deriving the formula that would be amazing!

[@p.t.9171]

My n is 230- so I can’t find the t, I’m stumped?

[@dmcguckian]

You need a t value for df= n-2 = 228. You can use software or you can use interpolation with a standard t table. This just means you approximate the t value using the two nearby t values by dividing the distance between the two surrounding t values by the difference between the degrees of freedom. For example, for df 200 and 95% confidence the t value is 1.972. For df 300 the value is 1.968. The difference between these values 1.972-1.968 is just .004. The difference between the degrees of freedom is 100 (300-200). This means it takes 100 degrees of freedom to change the t value by 0.004. The t value for Degrees of freedom 228 is therefore close to 1.972-.004/100*28 = 1.97088. This is not exactly correct, but it’s not too far from the correct value of 1.97