Question

In a sample of 200 households, the mean number of hours spent on social networking sites during the month of January was 55 hours. In a much larger study, the standard deviation was determined to be 7 hours. Assume the population standard deviation is the same. What is the 99% confidence interval for the mean hours devoted to social networking in January?

Answer 1

Answer:

53.7279 ≤ μ ≤ 56.2721

Step-by-step explanation:

If the mean of the sample is known, the population standard deviation is known and the size of the sample is bigger than 30, the (1-∝) confidence interval is calculated as:

$x-z_{\alpha/2}*\frac{s}{\sqrt{n}}$ ≤ μ ≤ $x+z_{\alpha/2}*\frac{s}{\sqrt{n}}$

Where x is the mean of the sample, s is the population standard deviation, n is the size of the sample and $z_{\alpha/2}$ is the value of the standard normal distribution in which:

P(Z>z)=∝/2

If 1-∝ is equal to 99%, ∝/2 is equal to 0.005 and $z_{\alpha/2}$ is equal to 2.57

So, if we replace x by 55, s by 7, n by 200 and $z_{\alpha/2}$ by 2.57, we get:

$55-2.57*\frac{7}{\sqrt{200}}$ ≤ μ ≤ $55+2.57*\frac{7}{\sqrt{200}}$

55 - 1.2721 ≤ μ ≤ 55 + 1.2721

53.7279 ≤ μ ≤ 56.2721

Answer 2

MOE = +\- 1.27

The 99% confidence interval ranges from 53.73 to 56.27 hours.