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a toy manufacturer wants to know how many new toys children buy each year. assume a previous study found the standard deviation to be 1.7 . she thinks the mean is 6.7 toys per year. what is the minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 98% level of confidence? round your answer up to the next integer.

User Diamondsea
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Final answer:

To ensure the estimate of the mean number of new toys children buy each year has an error of at most 0.14 at the 98% level of confidence, the minimum sample size required is 801, after rounding up the calculated sample size which is approximately 800.50721.

Step-by-step explanation:

To find the minimum sample size required to estimate the mean number of new toys children buy each year with a specified margin of error at a given level of confidence, we use the formula for the sample size of a mean:

n = (Z*(σ/E))^2

Where:

  • n is the sample size.
  • Z is the z-value corresponding to the desired level of confidence (in this case, 98%).
  • σ is the population standard deviation (in this case, 1.7).
  • E is the desired margin of error (in this case, 0.14).

Since the level of confidence is 98%, we find the z-value that corresponds to the upper 1% of the distribution because 100% - 98% = 2% and 2%/2 = 1% (as we consider both tails of the normal distribution).

The z-value from the z-tables for 98% confidence level (or upper 1%) is approximately 2.33. Now, we can calculate the sample size:

n = (2.33 * (1.7 / 0.14))^2

n = (2.33 * 12.143)^2

n = (28.2929)^2

n = 800.50721

Since we need a whole number for the sample size and we round up to ensure the margin of error is not exceeded, the minimum sample size required is 801.

User Cyague
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