Correct
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the $426 \mathrm{gram}$ setting. Based on a 13 bag sample where the mean is 433 grams and the standard deviation is 29 , is there sufficient evidence at the 0.05 level that the bags are overfilied? Assume the population distribution is approximately normal

Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.

Step 1 ：A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the $426 \mathrm{gram}$ setting. Based on a 13 bag sample where the mean is 433 grams and the standard deviation is 29 , is there sufficient evidence at the 0.05 level that the bags are overfilled? Assume the population distribution is approximately normal.

Step 2 ：To find the value of the test statistic, we need to use the formula for the z-score, which is \((\text{sample mean} - \text{population mean}) / (\text{standard deviation} / \sqrt{\text{sample size}})\). In this case, the sample mean is 433 grams, the population mean is 426 grams, the standard deviation is 29 grams, and the sample size is 13.

Step 3 ：Let's calculate the z-score: \[z = \frac{433 - 426}{29 / \sqrt{13}} \approx 0.870\]

Step 4 ：The calculated z-score is approximately 0.870. This value represents how many standard deviations the sample mean is away from the population mean. The positive value indicates that the sample mean is above the population mean.

Step 5 ：Final Answer: The value of the test statistic is approximately \(\boxed{0.870}\).