Quality Control at a Bottling Plant
A bottling plant fills 500 mL bottles. The fill volume is normally distributed with a mean of 500 mL and a standard deviation of 5 mL. The plant manager wants to ensure that at least 95% of bottles contain at least 495 mL. A sample of 30 bottles is taken hourly. The quality control inspector uses a 5% significance level to test whether the mean fill volume has decreased from 500 mL. In one hour, the sample mean is 497.5 mL. Assume the population standard deviation is known to be 5 mL.
1. State the null and alternative hypotheses and perform the hypothesis test. What conclusion should the inspector draw?
Read the worked answer
Null hypothesis ( H_0: \mu = 500 ) mL. Alternative hypothesis ( H_a: \mu < 500 ) mL (one-sided test). The test statistic is ( z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{497.5 - 500}{5/\sqrt{30}} = \frac{-2.5}{0.9129} \approx -2.74 ). The critical value for a one-sided test at ( \alpha = 0.05 ) is ( z_{0.05} = -1.645 ). Since ( -2.74 < -1.645 ), we reject ( H_0 ). There is sufficient evidence at the 5% level to conclude that the mean fill volume has decreased from 500 mL.
2. Based on the sample mean of 497.5 mL, what proportion of bottles would be expected to contain less than 495 mL? Does this meet the manager's requirement?
Read the worked answer
Using the sample mean as an estimate of the population mean, the proportion below 495 mL is ( P(X < 495) ) where ( X \sim N(497.5, 5^2) ). Standardize: ( z = \frac{495 - 497.5}{5} = -0.5 ). The proportion is ( \Phi(-0.5) \approx 0.3085 ), or 30.85%. The manager requires that at least 95% of bottles contain at least 495 mL, which means at most 5% can be below 495 mL. Since 30.85% > 5%, the requirement is not met. The process needs adjustment.
3. What sample size would be needed to detect a decrease of 2.5 mL with 90% power at the 5% significance level? Assume the population standard deviation remains 5 mL.
Read the worked answer
For a one-sided test, the required sample size is ( n = \left( \frac{(z_{\alpha} + z_{\beta}) \sigma}{\delta} \right)^2 ), where ( \delta = 2.5 ) mL, ( \sigma = 5 ) mL, ( z_{\alpha} = 1.645 ) (for ( \alpha = 0.05 ) one-sided), and ( z_{\beta} = 1.282 ) (for 90% power). Then ( n = \left( \frac{(1.645 + 1.282) \times 5}{2.5} \right)^2 = \left( \frac{2.927 \times 5}{2.5} \right)^2 = (2.927 \times 2)^2 = (5.854)^2 \approx 34.27 ). Round up to 35. So a sample size of 35 bottles is needed.