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.
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.
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