A company that produces detergents wants to estimate the mean amount of detergent in 64-ounce jugs at a 99% confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is .20 ounce. How large a sample should the company select so that the estimate is within .04 ounce of the population mean?
Use a formula to find sample size.
Here is one:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value = 2.58, sd = .20, E = .04, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
I'll let you take it from here.
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To determine the sample size needed to estimate the mean amount of detergent in 64-ounce jugs with a confidence level of 99% and a margin of error of .04 ounces, you can use the formula:
n = (Z * σ / E)²
Where:
- n is the required sample size.
- Z is the z-value corresponding to the desired confidence level. For a 99% confidence level, the z-value is 2.576.
- σ is the standard deviation of the population, which in this case is .20 ounce.
- E is the desired margin of error, which is .04 ounce.
Plugging in the values:
n = (2.576 * 0.20 / 0.04)²
Simplifying:
n = (12.88 / 0.04)²
n = 322.0²
n = 103,684
Therefore, the company should select a sample size of at least 103,684 to estimate the mean amount of detergent in 64-ounce jugs at a 99% confidence level with a margin of error of .04 ounce.
To determine the sample size needed, we can use the formula for sample size calculation for estimating population mean:
n = ((Z * σ) / E)²
Where:
- n is the sample size needed
- Z is the Z-score corresponding to the desired confidence level
- σ is the population standard deviation
- E is the desired margin of error
In this case, we want to estimate the mean amount of detergent in 64-ounce jugs with a 99% confidence level, and we want the estimate to be within 0.04 ounces of the population mean.
First, let's find the value of Z for a 99% confidence level. The Z-score associated with a 99% confidence level is approximately 2.576.
Next, we can substitute the values into the formula:
n = ((2.576 * 0.20) / 0.04)²
n = (0.5152 / 0.04)²
n = 12.88²
n ≈ 166.0544
Since we cannot have a fractional sample size, we need to round up to the nearest whole number. Therefore, the company should select a sample size of at least 167 jugs to estimate the mean amount of detergent with a 99% confidence level and a margin of error of 0.04 ounce.