Posted by **kim** on Tuesday, March 5, 2013 at 9:47pm.

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**, Wednesday, March 6, 2013 at 6:02pm
Try a one-sample t-test since the sample size is fairly small. This will be a one-tailed test (the alternate hypothesis will show a specific direction). Formula for a one-sample t-test 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 t-tables that show values for one-tailed and two-tailed 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 one-tailed test at .05 level of significance. Compare that value you find from the table to your t-test statistic calculated above. If the t-test statistic exceeds the critical value from the t-table, reject the null. If the t-test statistic does not exceed the critical value from the t-table, do not reject the null. Draw your conclusions from there.

I hope this helps.

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