Questions: A machine is set to fill paint cans with a mean of 121 ounces and a standard deviation of 0.30 ounce. A random sample of 45 cans has a mean of 120.9 ounces. The machine needs to be reset when the mean of a random sample is unusual. Does the machine need to be reset? Explain. it is that you would have randomly sampled 45 cans with a mean equal to 120.9 ounces, because it within the range of a usual event, namely within of the mean of the sample means.

Solution Steps

To determine if the sample mean is unusual, we first calculate the probability that the sample mean falls within 2 standard deviations of the population mean. The range for usual events is given by:

\[ \text{Range} = \left( \mu - 2 \cdot \frac{\sigma}{\sqrt{n}}, \mu + 2 \cdot \frac{\sigma}{\sqrt{n}} \right) \]

Substituting the values:

\[ \text{Range} = \left( 121 - 2 \cdot \frac{0.30}{\sqrt{45}}, 121 + 2 \cdot \frac{0.30}{\sqrt{45}} \right) \]

Calculating the probability:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.0) - \Phi(-2.0) = 0.9545 \]

Thus, the probability that the sample mean is within 2 standard deviations is:

\[ P = 0.9545 \]

Next, we perform a hypothesis test to check if the sample mean is significantly different from the population mean. The null hypothesis \(H_0\) states that the sample mean is equal to the population mean (\(\mu_0 = 121\)). The test statistic is calculated as follows:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.30}{\sqrt{45}} \approx 0.0447 \]

The Z-test statistic is then calculated using:

\[ Z = \frac{\bar{x} - \mu_0}{SE} = \frac{120.9 - 121}{0.0447} \approx -2.2361 \]

For a two-tailed test, the p-value is calculated as:

\[ P = 2 \times (1 - T(|z|)) \approx 0.0253 \]

Since the p-value \(0.0253\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis. This indicates that the sample mean is significantly different from the population mean.

Thus, we conclude that the machine needs to be reset because the sample mean is unusual.

Final Answer