In a goodness-of-fit 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

Degrees of freedom for a goodness-of-fit test is k-1 (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.

The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population

To find the critical value for the goodness-of-fit test, you can use the chi-square distribution.

First, let's determine the degrees of freedom (df) for the test. In this case, since you have 6 intervals used for testing, you have (6 - 1) = 5 degrees of freedom.

Next, you need to find the critical value from the chi-square distribution table, using the degrees of freedom (df) and the level of significance (α). In this case, the level of significance is 5%, which corresponds to an α of 0.05.

Consulting the chi-square distribution table, with 5 degrees of freedom and α = 0.05, the critical value is approximately 11.1433.

Therefore, your answer A) 11.1433 is correct.