Test 3
Question 10 of 20 (10 points) 1 Question Attempt 1 of 1
8
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$\equiv 17$
$=18$
$=19$
A potato chip company wants to evaluate the accuracy of its potato chip bag-filling machine. Bags are labeled as containing 8 ounces of potato chips. A simple random sample of 12 bags had mean weight 8.12 ounces with a sample standard deviation of 0.2 ounce. Assume the weights are approximately normally distributed. Construct a $90 \%$ confidence interval for the population mean weight of bags of potato chips. Round the answers to at least two decima places.
A $90 \%$ confidence interval for the mean weight of bags of potato chips is $\square< \mu< \square$.
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Submit A

Step 1 ：Given in the problem, we have: \(\bar{x} = 8.12\) ounces, \(s = 0.2\) ounce, and \(n = 12\) bags.

Step 2 ：The Z-score for a 90% confidence interval is approximately 1.645.

Step 3 ：Substitute these values into the formula for a confidence interval: \(CI = \bar{x} \pm Z * (s/\sqrt{n})\).

Step 4 ：Calculate the standard error (s/\sqrt{n}): \(SE = 0.2/\sqrt{12} \approx 0.057735\).

Step 5 ：Multiply the Z-score by the standard error: \(Z * SE = 1.645 * 0.057735 \approx 0.0949\).

Step 6 ：Add and subtract this value from the sample mean to get the confidence interval: \(CI = 8.12 - 0.0949 < \mu < 8.12 + 0.0949\).

Step 7 ：\(\boxed{8.0251 < \mu < 8.2149}\) is the 90% confidence interval for the mean weight of bags of potato chips.