A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. Based on a 24 bag sample whiere the mean is 415 grams and the variance is 225 , is there sufficient evidence at the 0.1 lievel that the bags are underfiled or overifled? 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.

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Step 1 ：A manufacturer of chocolate chips wants to verify if its bag filling machine is functioning correctly at the 418 gram setting. A sample of 24 bags is taken, with a mean weight of 415 grams and a variance of 225 grams. We are to determine if there is sufficient evidence at the 0.1 level that the bags are underfilled or overfilled, assuming the population distribution is approximately normal.

Step 2 ：First, we need to calculate the standard deviation. The standard deviation is the square root of the variance. Given that the variance is 225, the standard deviation is \( \sqrt{225} = 15 \) grams.

Step 3 ：Next, we calculate the z-score. The z-score is a measure of how many standard deviations an element is from the mean. It is calculated using the formula \( z = \frac{{X_{bar} - \mu}}{{\sigma / \sqrt{n}}} \), where \( X_{bar} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size. Substituting the given values into the formula, we get \( z = \frac{{415 - 418}}{{15 / \sqrt{24}}} \).

Step 4 ：Calculating the above expression, we find that the z-score is approximately -0.980.

Step 5 ：Final Answer: The value of the test statistic, rounded to three decimal places, is \(\boxed{-0.980}\).