Welcome! A random pattern of 27 observations from alarge population has an average of 22 and a normal deviation of four.8. Can we conclude at = zero.01 that the populationmean is enormously below 24? On account that were testing if the populace meanis beneath 24, the substitute hypothesis might be that the mean is not up to 24. The null speculation consequently will likely be thatthe population imply is bigger or equal to 24. The value level for this experiment zero.01. Now, it is a left-tailed scan so the criticalor rejection vicinity might be in the left tail of the distribution. Due to the fact that we only have the pattern regular deviation,and not the population, we will be able to conduct a t-scan.As a way to find the valuable price setting apart thecritical vicinity from the non-principal vicinity, we seek advice the t-distribution desk. Now the t distribution uses degrees of freedom,df, as seen right here. The degrees of freedom for a one pattern ttest for mean is n 1. Due to the fact that the pattern size in our illustration is 27,the levels of freedom will probably be 27 1 which offers 26. When you consider that had been conducting a one-tailed testand = zero.01 for our test, the corresponding crucial value can be this 2.479 right here. However considering it is a left-tailed scan, itwill truly be a poor 2.479. So we say we reject the null speculation ifthe t-statistic is less than -2.479. Now substituting the values into the method,we calculate the test statistic as -2.165. And as you’ll discover right here, the scan statisticis no longer not up to -2.479, or just isn’t within the rejection area, so we fail to reject thenull hypothesis. And for the reason that we can not reject the null hypothesis,we cannot help the alternative that says that the mean is not up to 24. So at alpha = 0.01 we can say there’s notenough proof to conclude that the populace imply is less than 24.And that concludes this tutorial. Thanks for watching..