We have to find the minimum sample size.
The margin of error that it is aimed is ±0.2 hours from the real mean.
The population standard deviation is 0.8 hours.
The desired level of confidence is 96%. This corresponds to a z-score of 2.054.
We can relate sample size with the given information as:
![\begin{gathered} e=(\sigma)/(√(n))\cdot z \\ √(n)=(\sigma)/(e)\cdot z \\ n=((\sigma\cdot z)/(e))^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y6q27p9u8ttgyipmha4yfhum504mo3bddu.png)
If we replace e = 0.2, σ = 0.8 and z = 2.054 we can calculate the sample size n as:
![\begin{gathered} n=((0.8*2.054)/(0.2))^2 \\ \\ n=(8.216)^2 \\ n\approx67.5 \\ n\approx68 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j2c4e3w4k1ycz7ba6wrzuoy1o2awwdk3hj.png)
Answer: the sample size has to be at least 68 people.
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