Question

An important part of any dispensing process is statistical quality control. machines are supposed to dispense on average 600 ml of soda to every glass. at a random point each hour, the owner dispenses and checks a glass to determine the actual volume dispensed. one day the volumes are: 600.15 599.92 599.85 599.92 599.81 600.14 600.04 599.98 is there sufficient evidence to conclude that the average volume of soda dispenses is different than 600 ml? test at alpha = 0.05. what are the critical values for the test? -2.365 ± 2.365 -1.895 ±1.96

Answer 1

The null and alternative hypotheses are:

`H_(0):\mu=600`

`H_(a):\mu \\eq 600`

The two tailed critical values at 0.05 significance level for df = 8 - 1 = 7 is found using t table and is given below:

`t_(critical)=\pm 2.365`

We can also use excel to find this critical value. The excel formula is:

=TINV(0.05,7)

Under null hypothesis, the test statistic is:

`t=\frac{\bar{x}-\mu}{(s)/(√(n)) }`

Where:

`\bar{x}=599.98` is the sample mean of given one day volumes.

`s=0.1259` is the sample standard deviation of given one day volumes.

`n=8` is the sample size

`\therefore, t=(599.98-600)/((0.1259)/(√(8)) )`

`=-0.45`

Conclusion: Since the test statistic does not lie outside the critical values, we therefore, fail to reject the null hypothesis and conclude that the average volume of soda dispenses is no different from 600 ml.