posted by K on .
average for all football teams was an 86 with variance of 16. the buckeyes program over the last 16 year was a 93. is there sufficient proof at the .05 level of significance to say buckeyes were higher than everyone else.
Use a one-sample z-test.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Note: Standard deviation is the square root of the variance.
Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.
I'll let you take it from here.