Posted by Stuart on Sunday, March 15, 2009 at 7:52pm.
In a goodnessoffit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 5% level of significance, the critical value for this test is:
A) 11.1433
B) 9.3484
C) 7.8147
D) 9.4877
I picked A 11.1433

Business Statistics  MathGuru, Sunday, March 15, 2009 at 11:14pm
Degrees of freedom for a goodnessoffit test is k1 (k = number of categories).
I'm not sure which is the correct response since I don't see the number of categories stated in the problem.