Question

300 high school students were asked how many hours of tv they watch per day. the mean was 2 hours, with a standard deviation of 0.5. using a 90% confidence level, calculate the maximum error of estimate. o 7.43% o 5.66% o 4.75% o 0.27%

Explanation:

Step1: Identify the z - score

For a 90% confidence level, the z - score $z_{\alpha/2}=1.645$.

Step2: Identify the standard deviation and sample size

The standard deviation $\sigma = 0.5$ and the sample size $n = 300$.

Step3: Calculate the maximum error of estimate

The formula for the maximum error of estimate $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Substitute the values: $E = 1.645\times\frac{0.5}{\sqrt{300}}\approx1.645\times\frac{0.5}{17.32}\approx1.645\times0.0289\approx0.0475$.

Answer:

None of the given options are correct. The maximum error of estimate is approximately 0.0475 or 4.75% (but the options seem to be percentage - formatted incorrectly in terms of the context of the problem). If we assume the options are just values representing the maximum error of estimate in decimal - converted to percentage, the correct option would be 4.75%.