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The population standard deviation for the weights of boxes of breakfast cereal is 0.6 ounces. If we want to be 
90% confident that the sample mean is within 1 ounce of the true population mean, what is the minimum sample size that can be taken?





z_(0.10)

z_(0.05)

z_(0.025)

z_(0.01)

z_(0.005)


1.282
1.645
1.960
2.326
2.576




Use the table above for the 
z-score, and be sure to round up to the next integer.
Provide your answer below:

The population standard deviation for the weights of boxes of breakfast cereal is 00.66 ounces. If we want to be 90% 90 \% confident that the sample mean is within 11 ounce of the true population mean, what is the minimum sample size that can be taken?\newline\begin{tabular}{|c|c|c|c|c|}\newline\hline z0.10 \mathbf{z}_{\mathbf{0 . 1 0}} & z0.05 \mathbf{z}_{\mathbf{0 . 0 5}} & z0.025 \mathbf{z}_{\mathbf{0 . 0 2 5}} & z0.01 \mathbf{z}_{\mathbf{0 . 0 1}} & z0.005 \mathbf{z}_{\mathbf{0 . 0 0 5}} \\\newline\hline 11.282282 & 11.645645 & 11.960960 & 22.326326 & 22.576576 \\\newline\hline\newline\end{tabular}\newlineUse the table above for the z z -score, and be sure to round up to the next integer.\newlineProvide your answer below:

Full solution

Q. The population standard deviation for the weights of boxes of breakfast cereal is 00.66 ounces. If we want to be 90% 90 \% confident that the sample mean is within 11 ounce of the true population mean, what is the minimum sample size that can be taken?\newline\begin{tabular}{|c|c|c|c|c|}\newline\hline z0.10 \mathbf{z}_{\mathbf{0 . 1 0}} & z0.05 \mathbf{z}_{\mathbf{0 . 0 5}} & z0.025 \mathbf{z}_{\mathbf{0 . 0 2 5}} & z0.01 \mathbf{z}_{\mathbf{0 . 0 1}} & z0.005 \mathbf{z}_{\mathbf{0 . 0 0 5}} \\\newline\hline 11.282282 & 11.645645 & 11.960960 & 22.326326 & 22.576576 \\\newline\hline\newline\end{tabular}\newlineUse the table above for the z z -score, and be sure to round up to the next integer.\newlineProvide your answer below:
  1. Identify Formula: Identify the formula to calculate the minimum sample size for a given confidence level and margin of error. The formula is:\newlinen=(Zσ/E)2n = (Z \cdot \sigma / E)^2\newlinewhere nn is the sample size, ZZ is the z-score corresponding to the desired confidence level, σ\sigma is the population standard deviation, and EE is the margin of error.
  2. Determine Z-Score: Determine the z-score that corresponds to a 90%90\% confidence level. Looking at the provided z-score table, the z-score that corresponds to a 90%90\% confidence level (which leaves 10%10\% in the tails, or 5%5\% in each tail) is 1.6451.645.
  3. Plug Values: Plug the values into the formula. We have σ=0.6\sigma = 0.6 ounces, E=1E = 1 ounce, and Z=1.645Z = 1.645. Now we can calculate the sample size:\newlinen=(1.645×0.6/1)2n = (1.645 \times 0.6 / 1)^2
  4. Perform Calculation: Perform the calculation: n=(0.987)2n = (0.987)^2 n=0.974169n = 0.974169
  5. Round Up: Since we cannot have a fraction of a sample, we need to round up to the next whole number. The minimum sample size is therefore 11 when rounded up.

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