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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 409 gram setting. It is believed that the machine is underfilling the bags. A 20 bag sample had a mean of 399 grams with a standard deviation of 21. Assume the population is normally distributed. A level of significance of 0.02 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

User Pwilcox
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The P-value for the one-sample t-test is approximately 0.0232. At a 0.02 significance level, we reject the null hypothesis.

To test whether the bag filling machine is underfilling at the 409 gram setting, we can use a one-sample t-test. The null hypothesis (H0) is that the mean filling weight is 409 grams, and the alternative hypothesis (H1) is that the mean is less than 409 grams.

1. Formulate Hypotheses:

- Null Hypothesis (H0): μ = 409 grams

- Alternative Hypothesis (H1): μ < 409 grams

2. Calculate the Test Statistic:

- The formula for the t-test statistic is:
\[ t = \frac{{\bar{X} - \mu}}{{s/√(n)}} \]

where
\(\bar{X}\)is the sample mean,
\(\mu\) is the population mean, s is the sample standard deviation, and n is the sample size.

Substituting the given values, we get
\[ t = \frac{{399 - 409}}{{21/√(20)}} \]

3. Determine the P-value:

- Using the t-distribution table or statistical software, find the P-value associated with the calculated t-statistic.

4. Make a Decision:

- Compare the P-value to the significance level (0.02). If P-value < 0.02, reject the null hypothesis.

For the P-value calculation, you can use statistical software or online calculators. The P-value is the probability of observing a sample mean as extreme as 399 grams (or more extreme) if the true population mean is 409 grams. Report the P-value as a decimal rounded to four decimal places or as an interval if needed.

User Andreszs
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