What sample size would you need to estimate the average amount of time spent on non-school related assignments, within $2$ minutes if you wanted to be $97 \%$ confident in your estimate, given $\sigma=9.5$ .

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Q. What sample size would you need to estimate the average amount of time spent on non-school related assignments, within

2

minutes if you wanted to be

97 \%

confident in your estimate, given

\sigma=9.5

Identify Formula: Identify the formula to calculate the sample size for estimating a population mean with a certain level of confidence. The formula is: $\newline$ $n = (Z \cdot \sigma / E)^2$ $\newline$ where $n$ is the sample size, $Z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error (the maximum allowable error in the estimate).
Determine Z-Score: Determine the z-score corresponding to a $97\%$ confidence level. This can be found using a z-table or a statistical calculator. For a $97\%$ confidence level, the z-score is approximately $2.17$ .
Plug Known Values: Plug the known values into the formula. We have $\sigma = 9.5$ (the standard deviation) and $E = 2$ (the desired margin of error). Now we can substitute these values into the formula: $\newline$ $n = (2.17 \times 9.5 / 2)^2$
Perform Calculations: Perform the calculations step by step. $\newline$ First, calculate the numerator of the fraction: $\newline$ $2.17 \times 9.5 = 20.615$ $\newline$ Next, divide by the margin of error, $E$ : $\newline$ $20.615 / 2 = 10.3075$ $\newline$ Finally, square the result to find $n$ : $\newline$ $10.3075^2 = 106.2445625$
Round Sample Size: Since the sample size must be a whole number, and you cannot have a fraction of a sample, round up to the nearest whole number. This ensures that the sample size is not smaller than needed for the desired confidence level. $\newline$ $n = 107$ (rounded up from $106.2445625$ )