Using the following information find a 95% confidence interval for the intercept in the regression of measured distance on recession velocity.
VARIABLE COEFFICIENT STANDARD ERROR
Constant 0.3391 0.1185
Velocity 0.001373 0.000227
T-STAT P-VALUE
3.369 0.0028
6.036 0.0000045
Est. of sigma = 0.4050 (22 d.f.)
We are understaffed in the statistics subject area, so are not able to answer most or all of your questions. I do apologize. We are alwys trying to find new good teachers to help
To find the 95% confidence interval for the intercept in the regression of measured distance on recession velocity, we need to use the coefficient and the standard error of the constant term.
The coefficient for the constant term is given as 0.3391, and the standard error is given as 0.1185.
Since we are interested in constructing a confidence interval at the 95% confidence level, we need to find the critical value associated with a 95% confidence level.
The critical value can be found using the t-distribution with the degrees of freedom mentioned (22 d.f.). For a 95% confidence interval, we need to find the critical value associated with a 2.5% significance level (since the t-distribution is symmetric).
Looking up the t-distribution table or using statistical software, we find that the critical value is approximately 2.074.
Next, we can calculate the margin of error by multiplying the critical value by the standard error:
Margin of Error = 2.074 x 0.1185 = 0.2459.
Finally, we can construct the 95% confidence interval by adding and subtracting the margin of error from the coefficient:
95% Confidence Interval = 0.3391 ± 0.2459 = (0.0932, 0.5849).
Therefore, the 95% confidence interval for the intercept in the regression of measured distance on recession velocity is (0.0932, 0.5849).