Posted by kim on .
A balloon mannufacturer kows that historically her company has produced 614 balloons per day (standard deviation =36) she has changed the process and wants to know wheathert he new process is increased the number of ballons made. the sample mean is 620. 0.5, n=16

statistics 
MathGuru,
Try a onesample ttest since the sample size is fairly small. This will be a onetailed test (the alternate hypothesis will show a specific direction). Formula for a onesample ttest is the following:
z = (sample mean  population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (620  614)/(36/√16) = ?
I'll let you finish the calculation.
There are ttables that show values for onetailed and twotailed tests. If n = 16, degrees of freedom would be n  1, which is 15. Check the critical value from the table using degrees of freedom for a onetailed test at .05 level of significance. Compare that value you find from the table to your ttest statistic calculated above. If the ttest statistic exceeds the critical value from the ttable, reject the null. If the ttest statistic does not exceed the critical value from the ttable, do not reject the null. Draw your conclusions from there.
I hope this helps.