Question

A producer of a variety of salty snacks would like to estimate the average weight of a bag of BBQ potato chips produced during the filling process at one of its plants. Determine the sample size needed to construct a 99​% confidence interval with a margin of error equal to 0.007 ounces. Assume the standard deviation for the potato chip filling process is 0.05 ounces.

The sample size needed is______________​(Round up to the nearest​ integer.)

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Determine the sample size needed to construct a 95%confidence interval to estimate the average GPA for the student population at a college with a margin of error equal to 0.2. Assume the standard deviation of the GPA for the student population is 2.5.

The sample size needed is________________​(Round up to the nearest​ integer.)

Answer 1

To construct a 99% confidence interval with 0.007 ounces margin of error, a sample size of 125 is needed. To construct a 95% confidence interval with a 0.2 margin of error, a sample size of 241 is needed.

Step-by-step explanation:

To determine the sample size needed to construct a 99% confidence interval with a margin of error equal to 0.007 ounces, we can use the formula:

Sample size = ((Z-value)^2 * (standard deviation)^2) / (margin of error)^2

Plugging in the given values:

Sample size = ((2.33)^2 * (0.05)^2) / (0.007)^2

Sample size ≈ 124.489

Rounding up to the nearest integer, the sample size needed is 125.

For the second question, to determine the sample size needed to construct a 95% confidence interval with a margin of error equal to 0.2, we can use the same formula:

Sample size = ((Z-value)^2 * (standard deviation)^2) / (margin of error)^2

Plugging in the given values:

Sample size = ((1.96)^2 * (2.5)^2) / (0.2)^2

Sample size ≈ 240.1

Rounding up to the nearest integer, the sample size needed is 241.