Question

You are in charge of quality control at the Utica Ketchup Bottling plant. Your bottles are labeled 14 ounces but due to natural imperfections in the bottling process, not every bottle contains precisely the same amount of ketchup. Let’s assume that the volumes are distributed Normally with a standard deviation of 0.1 oz. Before a shipment of ketchup goes out the door, you take a random sample of 30 bottles and measure the volume of each. Before you approve a shipment, you must be convinced that the mean volume of the shipment exceeds 14 ounces with a significance level of α = 0.05. Your random sample of 30 ketchup bottles has average volume ¯x = 14.035 oz. (Please only answer one part at a time to give other students a chance to answer as well! Start with the first one!)

1. Carefully state the null and alternative hypotheses for this scenario. Is this a one-sided or two-sided test?

2. If you want to show that the average volume is at least 14 ounces and the average volume of your sample is 14.035 ounces, why can you not immediately approve the shipment based on the fact that 14.035 > 14 with no further testing?

Answer 1

Data given is:

Sample size, n = 30

Sample mean, m = 14.035

Standard deviation, S = 0.1

We conduct a one sided hypothesis test here, with the following hypotheses:

H0: = 14

Ha: > 14

(b)

We can't directly reach to the conclusion that mean weight for the whole population is greater than 14, just based on the observation that sample mean is 14.035,which is greater than 14. This is because it can possibly be the base that we obtained such a sample just out of pure luck. So in order to eradicate the luck factor and to be sure that the whole population of bottles indeed has a mean weight atleast 14 ounces, we need to confirm this by conducting a hypothesis test and finding the p-value for the sample.

First we calculate standard error:

SE = S/(n^0.5) = 0.1/(30^0.5) = 0.0183

Next we calculate test statistic:

z = (m-14)/SE = (14.035-14)/0.0183 = 1.913

The corresponding p-value for this z-score is obtained from the z-table as:

p = 0.028

Given significance level, a = 0.05

Since p < a, we have to reject the null hypothesis here.

Thus NOW WE CAN conclude from the given data that mean weight of bottles is atleast 14 ounces.